Top Open Problems in Math by LLM-Assessed Importance

ProofAtlas published a Top 500 ranking of open mathematical problems ordered by LLM-assessed importance, with P versus NP ranked first. The ranking was produced by pairwise comparisons between problems combined through a reliability-weighted model using family weights of OpenAI 1.00, Claude 1.00, GLM 0.95, and DeepSeek 0.90, and ProofAtlas states it is a model-based assessment rather than expert consensus or a forecast of which problems will be solved next. Uncertainty ranges and comparison counts derive from a preceding fit of 1,275 targets, while the current eligible pool holds 1,273 and separately dated status notes cover 144 of the 500 entries.

The ProofAtlas Top 500 500 mathematical problems ranked by LLM-assessed importance Explore the questions shaping mathematics, computation, and mathematical physics. This is a model-based assessment of mathematical importance, not an expert consensus or a forecast of which problems will be solved next. Resolved problems from website and source editions resolved-problems remain in the history with their original ranks and sources. This edition retains 2 entries on reviewed holds; their ranks are conditional. Each entry shows its recorded reason. How the ranking works What is assessed. Each entry is a specific mathematical question. Importance considers the significance of a resolution, centrality within its field, reach across fields, scholarly and historical fame, broader recognition, and scientific or practical impact. How the order is formed. LLMs compare pairs of problems. A reliability-weighted model combines those judgments into the official ranking, with calibration across model families. The family weights are OpenAI 1.00, Claude 1.00, GLM 0.95, and DeepSeek 0.90. These are modeling choices, not measured probabilities of correctness. How to read a position. Nearby ranks can be uncertain. The expanded entries include available ranking ranges and comparison counts. Filtering or alphabetizing the directory preserves every problem’s official rank. Diagnostics from the preceding source fit. After the reported resolutions, official ranks were compacted without a new numerical fit. The displayed uncertainty ranges and comparison counts come from the preceding fit of 1,275 targets; the current eligible pool contains 1,273. These diagnostics were not recomputed for the smaller pool. Related and nested problems. A substantial subproblem or generalization may receive its own rank when it has independent mathematical standing. Families identify narrow groups of related questions; an implication alone does not merge their families. The ranking process checks family crowding, including a diagnostic retaining the highest-ranked member of each family. Family membership does not itself change scores, impose a quota, or create an automatic penalty. Dates and sources. The edition date identifies this list. A status review date, when recorded, appears separately inside each entry; it does not promise that the literature has been checked since then. Source links identify the exact formulation and available status evidence. The ranking preserves a fixed edition. Separately dated status notes cover 144 of 500 entries; this is not a fresh literature review of the whole list. Recent resolution claims and formulation concerns are marked below with their historical ranks retained. Find an open problem Open a row for its exact statement, significance, status, sources, and related research. Families are ordered by their highest-ranked member in the complete edition. Families have no importance score; every problem keeps its official rank. Filters show matching members within that fixed family order. No Top 500 problems match. Try a broader phrase or clear the filters. Resolved problems matching your search These entries remain in the status history, outside the open ranking. Matching problems outside the current Top 500 These problems remain in the full ranking. They are not included in the count above. 1. Rank 001 P versus NP When a yes-or-no problem has short evidence for each yes answer that can be checked in polynomial time, must a deterministic computer also be able to solve the problem in polynomial time? Here polynomial time measures how the work grows with the input size.Theoretical computer science Classical complexity separations · 5 problems family-context-problem-p-versus-np Exact statementDetermine whether P = NP, that is, whether every decision problem whose solutions can be verified in polynomial time by a deterministic Turing machine can also be solved in polynomial time by a deterministic Turing machine. Why it mattersEquality would give polynomial-time algorithms for every NP decision problem; separation would establish that some efficiently checkable problems resist all such algorithms. This addresses computational efficiency as input size grows. Status Open P versus NP remains open on the maintained Clay statement and the bounded claim sweep; no accepted exact-target resolution was established. Status reviewed Sources - Clay Mathematics Institute, The Millennium Prize Problems. https://www.claymath.org/library/monographs/MPPc.pdf Statement source - P vs NP https://www.claymath.org/millennium/p-vs-np/ Status source · Retrieved Sep 14, 2026 - Constructive solvability and the P versus NP problem — Arne Hole https://arxiv.org/abs/2406.16843 Status source · Retrieved Sep 14, 2026 - PNP Labs — P versus NP formal reconstruction https://pnplabs.com.au/ Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyClassical complexity separations · 5 problems in this edition. Family membership alone does not assert an implication. Ranking uncertainty90% source-fit bootstrap rank range: 1–1. Source-fit opponent count: 34. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 2. Rank 002 Riemann Hypothesis Do all zeros of the Riemann zeta function inside the strip with real part between zero and one lie exactly on its middle line?Number theory & arithmetic geometry Riemann hypotheses and exceptional zeros · 4 problems family-context-problem-riemann-hypothesis Exact statementFor the meromorphic continuation of the Riemann zeta function zeta s =sum {n =1} n^ -s , initially defined for Re s 1, every zero s in the open critical strip 0<Re s <1 has Re s =1/2. This is the Riemann Hypothesis for zeta alone; it does not assert simplicity of its zeros. Why it mattersA proof would sharply control fluctuations in the distribution of primes: the prime-counting function would differ from its logarithmic-integral approximation by at most O √x log x . A disproof would invalidate this equivalent error bound. Status Open Riemann remains open. September 2026 progress on the proportion of simple critical-line zeros is partial, not proof that every nontrivial zero lies on the line. Uncorroborated or inaccessible proof claims were not accepted. Status reviewed Sources - Enrico Bombieri, Problems of the Millennium: the Riemann Hypothesis, Section I and the prime-counting equivalence. https://www.claymath.org/wp-content/uploads/2022/05/riemann.pdf Statement source - Riemann Hypothesis https://www.claymath.org/millennium/riemann-hypothesis/ Status source · Retrieved Sep 14, 2026 - A new proof that more than 2/3 of the zeros of the Riemann zeta function are simple and on the critical line https://arxiv.org/abs/2609.02882 Status source · Retrieved Sep 14, 2026 - Lamzouri lecture at Max Planck Institute for Mathematics, September 3, 2026 https://math-events.uni-bonn.de/event/1560/ Status source · Retrieved Sep 14, 2026 - On the Extended, Generalized, and Grand Riemann Hypotheses: A Unified Approach Based on the General Properties of L-Functions — Weicun Zhang, v19 https://www.preprints.org/manuscript/202506.0481 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyRiemann hypotheses and exceptional zeros · 4 problems in this edition. Family membership alone does not assert an implication. Ranking uncertainty90% source-fit bootstrap rank range: 2–2. Source-fit opponent count: 53. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 3. Rank 003 Yang-Mills Existence and Mass Gap Can a nontrivial Yang–Mills quantum field theory in four-dimensional spacetime be constructed rigorously for every compact simple gauge group—the choice of internal symmetry—with a positive, finite gap between vacuum energy and the remaining energy spectrum?Mathematical physics Exact statementFor every compact simple gauge group G, prove that a nontrivial four-dimensional quantum Yang-Mills theory on R^4 exists with all existence requirements of Jaffe-Witten, Quantum Yang-Mills Theory, sections 3-4, including axiomatic properties at least as strong as their cited references 45,35 , the required gauge-invariant local fields and short-distance compatibility specified there. Its vacuum Hamiltonian H has no spectrum in 0,Delta for some Delta 0, and the supremum of such Delta is finite. The official reconstruction and existence requirements are retained by reference, not replaced by an unrestricted equivalence of arbitrary Euclidean and Minkowski theories. An isolated first excited eigenstate, confinement and asymptotic completeness are not additional requirements. Why it mattersThe construction would turn the specified four-dimensional quantum gauge-field model into a theory satisfying the required mathematical axioms. Its spectrum would have a positive, finite gap above vacuum energy, without requiring an isolated first excited eigenstate. Status Open The universal four-dimensional Yang–Mills existence/mass-gap target remains open. Three-dimensional outlines, individual gauge-group claims and withdrawn postings do not establish the full target. Status reviewed Sources - Arthur Jaffe and Edward Witten, Quantum Yang-Mills Theory, official Clay problem description. https://www.claymath.org/wp-content/uploads/2022/06/yangmills.pdf Statement source - Yang-Mills & the Mass Gap https://www.claymath.org/millennium/yang-mills-the-maths-gap/ Status source · Retrieved Sep 14, 2026 - Quantum Yang-Mills Theory — Jaffe and Witten https://www.claymath.org/wp-content/uploads/2022/06/yangmills.pdf Status source · Retrieved Sep 14, 2026 - Towards a Proof of Mass Gap in 3d Yang-Mills Theory — V. P. Nair https://arxiv.org/abs/2608.10133 Status source · Retrieved Sep 14, 2026 - A Constructive Proof of Existence and Mass Gap for Pure SU 3 Yang-Mills in Four-Dimensional Space-Time — D. C. Jacobsen https://arxiv.org/abs/2506.00284 Status source · Retrieved Sep 14, 2026 Problem familyYang-Mills Existence and Mass Gap · 1 problem in this edition. Family membership alone does not assert an implication. - 3 · Yang-Mills Existence and Mass Gap this problem Ranking uncertainty90% source-fit bootstrap rank range: 3–5. Source-fit opponent count: 47. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 4. Rank 004 Hodge Conjecture Hodge theory singles out certain rational cohomology classes, which encode topological features of a smooth complex projective variety. Must every such class be a rational combination of classes represented by algebraic subvarieties?Geometry & topology Hodge conjectures · 3 problems family-context-problem-hodge-conjecture Exact statementDetermine whether every rational Hodge class on a smooth complex projective algebraic variety is a rational linear combination of classes of algebraic subvarieties algebraic cycles . Why it mattersCohomology records features of a variety, while algebraic cycles are built from its algebraic subvarieties. A proof would show that every rational class of the specified Hodge type can be accounted for by those algebraic pieces. Status Open The rational Hodge conjecture remains open. Conditional witness axioms, selected Fermat varieties and inaccessible full-proof claims do not establish the all-varieties target. Status reviewed Sources - Clay Mathematics Institute, The Millennium Prize Problems. https://www.claymath.org/library/monographs/MPPc.pdf Statement source - Hodge Conjecture https://www.claymath.org/millennium/hodge-conjecture/ Status source · Retrieved Sep 14, 2026 - A Proof of the Hodge Conjecture Derived from One Semiregular Witness Axiom — Mohammad F Islam https://zenodo.org/records/22701510 Status source · Retrieved Sep 14, 2026 - The Hodge conjecture for Fermat fourfolds of odd degree at most 199 — Rifat Jumagulov https://arxiv.org/abs/2608.18134 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyHodge conjectures · 3 problems in this edition. Family membership alone does not assert an implication. - 4 · Hodge Conjecture this problem - 24 · Grothendiecks Generalized Hodge Conjecture problem-grothendiecks-generalized-hodge-conjecture - 90 · Deligne's Absolute Hodge Conjecture problem-top500-omission-w045049-absolute-hodge-conjecture Ranking uncertainty90% source-fit bootstrap rank range: 3–7. Source-fit opponent count: 32. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 5. Rank 005 Birch and Swinnerton-Dyer Conjecture For an elliptic curve over the rational numbers, its rank counts the independent infinite-order generators of its group of rational points. Does this rank equal the order of vanishing of its L-function at s = 1—the multiplicity of the zero there, taken as zero when the function is nonzero?Number theory & arithmetic geometry Elliptic curve ranks and congruent numbers · 3 problems family-context-problem-birch-and-swinnerton-dyer-conjecture Exact statementProve or disprove that, for every elliptic curve E over the rational numbers, the rank of the finitely generated abelian group E Q of rational points equals the order of vanishing at s = 1 of the associated Hasse–Weil L-function L E, s . Why it mattersA proof would identify an algebraic measure of the curve’s rational points with an analytic measurement of its L-function. A counterexample would locate a curve where those two measurements differ. Status Open BSD rank equality remains open. The located all-rank working-draft claim lacks independent acceptance; recent uses and special cases do not supply a universal proof. Status reviewed Sources - Clay Mathematics Institute, The Millennium Prize Problems. https://www.claymath.org/library/monographs/MPPc.pdf Statement source - Birch and Swinnerton-Dyer Conjecture https://www.claymath.org/millennium/birch-and-swinnerton-dyer-conjecture/ Status source · Retrieved Sep 14, 2026 - Introduction to the Birch and Swinnerton-Dyer Conjecture — Fabio Ferrari Ruffino https://arxiv.org/abs/2608.25975 Status source · Retrieved Sep 14, 2026 - Proof of the Birch and Swinnerton-Dyer Conjecture — Yoshinori Shimizu https://zenodo.org/records/17010555 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyElliptic curve ranks and congruent numbers · 3 problems in this edition. Family membership alone does not assert an implication. - 5 · Birch and Swinnerton-Dyer Conjecture this problem - 150 · Congruent Number Problem problem-congruent-number-problem - 194 · Parity conjecture for elliptic curves problem-parity-conjecture-for-elliptic-curves Ranking uncertainty90% source-fit bootstrap rank range: 4–8. Source-fit opponent count: 31. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 6. Rank 007 Global Langlands Functoriality Conjecture Langlands L-groups encode symmetries used to organize automorphic representations. For quasisplit reductive groups over number fields, does every admissible map between their L-groups transfer these representations while matching their local arithmetic data at all but finitely many places?Number theory & arithmetic geometry Langlands functoriality and modularity · 2 problems family-context-problem-global-langlands-functoriality-conjecture Exact statementFor every number field F, quasisplit connected reductive F-groups G-prime and G, and admissible L-homomorphism rho:L G-prime - L G over Gamma F, commuting with its projections, in Arthur Principle1.1's convention , every automorphic representation pi-prime of G-prime A F , in the inclusive Langlands sense cited there, has an automorphic representation pi of G A F whose unramified Satake classes equal the images under rho v of those of pi-prime outside a finite set of places. This is existence of a weak transfer, not uniqueness or a prescribed transfer function. Why it mattersA proof would guarantee the stated transfer across every group map in scope, linking automorphic representations through their local parameters. The existence claim permits more than one transfer and leaves its exceptional places finite. Status Open · subcases solved Global functoriality remains open, with solved subcases. Existing restricted transfers and recent conjectural use do not establish the full specified transfer theorem; evidence of its current general status remains limited. Status reviewed Sources - Classifying automorphic representations, Clay collected-works item76 https://www.claymath.org/library/cw/arthur/pdf/76.pdf Statement source - Functoriality and the Trace Formula — James Arthur https://www.math.toronto.edu/arthur/pdf/arthur-func-and-trace-formula.pdf Status source · Retrieved Sep 14, 2026 - Math 590: Open Problems in Number Theory, Spring 2026 — Duke University https://sites.math.duke.edu/~pierce/documents/Math590.pdf Status source · Retrieved Sep 14, 2026 Problem familyLanglands functoriality and modularity · 2 problems in this edition. Family membership alone does not assert an implication. - 7 · Global Langlands Functoriality Conjecture this problem - 39 · Modularity of Elliptic Curves over Number Fields End K E =Z problem-modularity-of-elliptic-curves-over-number-fields Ranking uncertainty90% source-fit bootstrap rank range: 4–8. Source-fit opponent count: 39. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 7. Rank 008 Existence of One-Way Functions Do one-way functions exist: functions that a deterministic computer can evaluate in polynomial time, but that every randomized polynomial-time algorithm reverses on a random input’s output with success eventually smaller than every inverse power of the input length? Reversing means finding any input with that output.Cryptography, coding, information & optimization One-way functions and permutations · 2 problems family-context-problem-existence-of-one-way-functions Exact statementProve or disprove that there exists a deterministic polynomial-time computable function family f n:{0,1}^n- {0,1}^{poly n } such that every probabilistic polynomial-time algorithm inverts f n on a uniformly random input with only negligible probability. Why it mattersA construction would unconditionally separate efficient forward computation from hard inversion on random inputs, even when the inverter uses randomness. Every randomized polynomial-time inverter would have only negligible success probability as input length grows. Status Open Unconditional existence of one-way functions remains open. Conditional constructions and hardness characterizations do not constitute an unconditional solved subcase of this existential target. Status reviewed Sources - Shuichi Hirahara, Zhenjian Lu, and Igor C. Oliveira, One-Way Functions and pKt Complexity, IACR ePrint 2024/1388 2024 . https://eprint.iacr.org/2024/1388.pdf Statement source - One-Way Functions and pKt Complexity — Hirahara, Lu and Oliveira https://wrap.warwick.ac.uk/id/eprint/187935/1/WRAP-One-way-functions-pKt-complexity-24.pdf Status source · Retrieved Sep 14, 2026 - One-Way Functions and Boundary Hardness of Randomized Time-Bounded Kolmogorov Complexity — Liu and Pass https://drops.dagstuhl.de/entities/document/10.4230/LIPIcs.ITCS.2026.97 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyOne-way functions and permutations · 2 problems in this edition. Family membership alone does not assert an implication. - 8 · Existence of One-Way Functions this problem - 57 · Existence of One-Way Permutations problem-existence-of-one-way-permutations Ranking uncertainty90% source-fit bootstrap rank range: 5–8. Source-fit opponent count: 31. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 8. Rank 009 The abc Conjecture of Masser and Oesterle Suppose coprime positive integers satisfy a + b = c, and R is the product of the distinct primes dividing abc. For every positive ε, is c less than C ε times R^ 1 + ε , with one constant C ε working for every such triple?Number theory & arithmetic geometry abc and Szpiro conjectures · 2 problems family-context-problem-the-abc-conjecture-of-masser-and-oesterle Open · disputed claim Exact statementProve that for every epsilon 0, there exists a constant C epsilon such that for all coprime positive integers a + b = c, c < C epsilon rad abc ^{1+epsilon}. Why it mattersA proof would uniformly restrict how large an additive triple can be relative to its distinct prime factors. A disproof would find a positive epsilon for which no constant controls all such triples. Status Open · disputed claim abc retains open disputed claim. Joshi's February 2025 revision is not adjudicated by the older Scholze–Stix objection to Mochizuki; the July 2026 abc-type proposal is another conjecture. Current acceptance of a full proof was not established. Status reviewed Sources - Distance Between Cubics and Rationals, Results in Mathematics 2026 , statement of the abc conjecture https://link.springer.com/article/10.1007/s00025-026-02616-5 Statement source - Construction of Arithmetic Teichmuller Spaces IV: Proof of the abc-conjecture — Kirti Joshi https://arxiv.org/abs/2403.10430 Status source · Retrieved Sep 14, 2026 - Why abc is still a conjecture — Peter Scholze and Jakob Stix, July 16, 2018 https://www.math.uni-bonn.de/people/scholze/WhyABCisStillaConjecture.pdf Status source · Retrieved Sep 14, 2026 - The abc Conjecture Revisited — Patrick Letendre, July 2026 https://arxiv.org/abs/2607.07641 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyabc and Szpiro conjectures · 2 problems in this edition. Family membership alone does not assert an implication. - 9 · The abc Conjecture of Masser and Oesterle this problem - 101 · Classical Szpiro Discriminant–Conductor Conjecture over Q problem-top500-omission-w050054-classical-szpiro-discriminantconductor-conjecture-over-q Ranking uncertainty90% source-fit bootstrap rank range: 9–14. Source-fit opponent count: 31. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 9. Rank 010 Generalized Riemann Hypothesis for Dirichlet L-functions Dirichlet L-functions extend the zeta-function setting by using periodic arithmetic weights. For every Dirichlet character, do all zeros of its L-function with real part between zero and one lie on the middle line, with real part one-half?Number theory & arithmetic geometry Riemann hypotheses and exceptional zeros · 4 problems family-context-problem-the-generalized-riemann-hypothesis-for-dirichlet-l-functions Exact statementFor every integer q =1 and every Dirichlet character chi modulo q, including principal and imprimitive characters, every zero s of the meromorphic continuation of L s,chi =sum {n =1}chi n n^ -s in 0<Re s <1 satisfies Re s =1/2. The continuation is from Re s 1. Only open-strip zeros are quantified: principal-character poles and boundary zeros of imprimitive Euler factors are not counterexamples. No simplicity or central nonvanishing requirement is added. Why it mattersA proof would bound how far one must search for a prime in an arithmetic progression. For q 3 and a coprime to q, the least prime congruent to a modulo q would be at most φ q log q ², where φ q counts the residue classes coprime to q. Status Open Dirichlet GRH remains open. No independently established full-target resolution was obtained; neighboring Riemann-family claims and source-access limitations remain qualified. Status reviewed Sources - NIST DLMF25.15, Dirichlet L-functions, equations25.15.1–4 and zeros; GRH convention in Kandhil–Languasco–Moree, Mathematische Annalen394:43 2026 . https://dlmf.nist.gov/25.15 Statement source - Large values of quadratic character sums — Dong, Wang, Wang, Zhang and Zhao https://arxiv.org/abs/2608.15773 Status source · Retrieved Sep 14, 2026 - On the Extended, Generalized, and Grand Riemann Hypotheses: A Unified Approach Based on the General Properties of L-Functions — Weicun Zhang, v19 https://www.preprints.org/manuscript/202506.0481 Status source · Retrieved Sep 14, 2026 Problem familyRiemann hypotheses and exceptional zeros · 4 problems in this edition. Family membership alone does not assert an implication. - 2 · Riemann Hypothesis problem-riemann-hypothesis - 10 · Generalized Riemann Hypothesis for Dirichlet L-functions this problem - 11 · Extended Riemann hypothesis for Dedekind zeta functions problem-extended-riemann-hypothesis-for-dedekind-zeta-functions - 48 · Nonexistence of Landau–Siegel zeros for quadratic Dirichlet L-functions problem-nonexistence-of-landau-siegel-zeros-for-quadratic-dirichlet-l-functions Ranking uncertainty90% source-fit bootstrap rank range: 9–14. Source-fit opponent count: 49. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 10. Rank 011 Extended Riemann hypothesis for Dedekind zeta functions Every number field—a finite algebraic extension of the rationals—has a Dedekind zeta function. Must all its zeros with real part between zero and one lie on the middle line, with real part one-half?Number theory & arithmetic geometry Riemann hypotheses and exceptional zeros · 4 problems family-context-problem-extended-riemann-hypothesis-for-dedekind-zeta-functions Open · disputed claim Exact statementFor every algebraic number field K a finite extension of Q, including K=Q , does every zero s of its Dedekind zeta function ζ K s in the critical strip 0<Re s <1 satisfy Re s =1/2? Why it mattersA proof would supply the zeta-function hypothesis used in effective Chebotarev estimates for how primes split in number fields, with their other conditions retained. Including the rational field would also establish the ordinary Riemann Hypothesis. Status Open · disputed claim Dedekind ERH remains open, with a disputed proof claim. Zhang's posting was revised to v19 on September 14, 2026; objections or prior assessments of v18 do not review v19. Independent acceptance of the newest claim was not established. Status reviewed Sources - American Institute of Mathematics RH project, The Extended Riemann Hypothesis; undated article18a https://www.aimath.org/WWN/rh/articles/html/18a/ Statement source - A random polynomial with multiplicative coefficients is almost surely irreducible — Varjú and Xu https://academic.oup.com/imrn/article/2026/16/rnag178/8761102 Status source · Retrieved Sep 14, 2026 - On the Extended, Generalized, and Grand Riemann Hypotheses: A Unified Approach Based on the General Properties of L-Functions — Weicun Zhang, v19 https://www.preprints.org/manuscript/202506.0481 Status source · Retrieved Sep 14, 2026 - Proof of generalized Riemann hypothesis for Dedekind zetas and Dirichlet L-functions — Andrzej Mądrecki https://arxiv.org/abs/0706.0256 Status source · Retrieved Sep 14, 2026 Problem familyRiemann hypotheses and exceptional zeros · 4 problems in this edition. Family membership alone does not assert an implication. - 2 · Riemann Hypothesis problem-riemann-hypothesis - 10 · Generalized Riemann Hypothesis for Dirichlet L-functions problem-the-generalized-riemann-hypothesis-for-dirichlet-l-functions - 11 · Extended Riemann hypothesis for Dedekind zeta functions this problem - 48 · Nonexistence of Landau–Siegel zeros for quadratic Dirichlet L-functions problem-nonexistence-of-landau-siegel-zeros-for-quadratic-dirichlet-l-functions Ranking uncertainty90% source-fit bootstrap rank range: 9–14. Source-fit opponent count: 41. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 11. Rank 012 NP versus P/poly Can every yes-or-no problem with short, efficiently checkable evidence for yes answers be solved by Boolean circuits of polynomial size, if a separate circuit may be chosen for each input length? No efficient method of constructing those circuits is required.Theoretical computer science Exact statementDetermine the truth of NP subseteq P/poly. Here NP consists of languages over finite binary strings accepted by nondeterministic Turing machines in polynomial time. P/poly consists of languages L for which there exist constants C =1 and integer k =1 and Boolean circuits C n for all n =0, of size at most C n+1 ^k, deciding membership in L on every length-n input. Circuits are finite acyclic single-output circuits with fan-in-two AND/OR, fan-in-one NOT, input bits and hardwired constants0/1; count all nodes. Each language has its own bound and circuit family, with no uniform generation requirement. Why it mattersContainment would collapse the polynomial hierarchy to its second level by the Karp–Lipton theorem. Noncontainment would establish a general circuit-size barrier for an NP language and imply P ≠ NP. Status Open NP containment in nonuniform P/poly remains open; February 2026 notes pose the matching general-circuit question. Restricted circuits and Karp–Lipton implications do not decide it. Status reviewed Sources - Arora and Barak, Computational Complexity: A Modern Approach, web draft January8,2007, Definitions6.1–6.3 and Theorem6.13; Aaronson, P ?= NP, Conjecture25. https://theory.cs.princeton.edu/complexity/book.pdf Statement source - Boolean Circuits and P/poly, Cornell CS 6810, February 26, 2026 https://www.cs.cornell.edu/courses/cs6810/2026sp/lec10.pdf Status source · Retrieved Sep 14, 2026 Problem familyNP versus P/poly · 1 problem in this edition. Family membership alone does not assert an implication. - 12 · NP versus P/poly this problem Ranking uncertainty90% source-fit bootstrap rank range: 9–14. Source-fit opponent count: 30. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 12. Rank 013 NP versus coNP Problem Whenever yes answers to a decision problem have polynomial-length certificates checkable in polynomial time, do its no answers always have such certificates as well?Theoretical computer science Classical complexity separations · 5 problems family-context-problem-np-versus-conp-problem Exact statementDetermine whether NP = coNP, that is, whether the class NP is closed under complementation. Equivalently, by the Cook-Reckhow criterion, decide whether there exists a polynomially bounded propositional proof system. Why it mattersEquality would make NP closed under exchanging yes and no answers and provide a polynomially bounded propositional proof system. Separation would rule out every such proof system. Status Open NP versus coNP remains open with an explicit claim warning. Lin's August 2026 v28 and Gordeev–Haeusler's May 2026 revision make opposing claims; older independent critiques do not automatically adjudicate those revisions. Restricted proof-system lower bounds are not the full separation. Status reviewed Sources - Scott Aaronson, "P =? NP," in Open Problems in Mathematics J. F. Nash Jr., M. Th. Rassias, eds. , Springer, 2016, pp. 1-122. https://www.scottaaronson.com/papers/pnp.pdf Statement source - Proof Complexity and Feasible Interpolation — Amirhossein Akbar Tabatabai 2025 https://arxiv.org/html/2505.03002v1 Status source · Retrieved Sep 14, 2026 - Superpolynomial Length Lower Bounds for Tree-Like Semantic Proof Systems with Bounded Line Size — de Rezende, Engström, Ghannane and Risse 2026 https://eccc.weizmann.ac.il/report/2026/078/ Status source · Retrieved Sep 14, 2026 - Simulating Polynomial-Time Nondeterministic Turing Machines via Nondeterministic Turing Machines — Tianrong Lin, v28 https://arxiv.org/abs/2406.10476 Status source · Retrieved Sep 14, 2026 - A Critique of Lin's On NP versus coNP and Frege Systems — DeJesse, Lyudovyk, Pai and Reidy 2025 https://arxiv.org/abs/2505.05658 Status source · Retrieved Sep 14, 2026 - Proofs of NP = coNP = PSPACE: Current upgrade — Lev Gordeev and Edward Hermann Haeusler, v3 https://arxiv.org/abs/2311.17939 Status source · Retrieved Sep 14, 2026 - A simplified lower bound for implicational logic — Emil Jeřábek September 2024 version; published 2025 https://arxiv.org/html/2303.15090v3 Status source · Retrieved Sep 14, 2026 Problem familyClassical complexity separations · 5 problems in this edition. Family membership alone does not assert an implication. - 1 · P versus NP problem-p-versus-np - 13 · NP versus coNP Problem this problem - 14 · P versus PSPACE problem-p-versus-pspace - 34 · P versus NP ∩ coNP problem-top500-omission-w050054-p-versus-np-conp - 103 · Superpolynomial Lower Bounds for Frege Propositional Proof Systems problem-superpolynomial-lower-bounds-for-frege-propositional-proof-systems Ranking uncertainty90% source-fit bootstrap rank range: 10–14. Source-fit opponent count: 29. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 13. Rank 014 P versus PSPACE Can every yes-or-no problem solvable by a deterministic computer using polynomially bounded memory also be solved deterministically in time polynomial in the input length?Theoretical computer science Classical complexity separations · 5 problems family-context-problem-p-versus-pspace Exact statementDetermine whether P = PSPACE, that is, whether every language decidable by a deterministic Turing machine in polynomial space is decidable by a deterministic Turing machine in polynomial time. Why it mattersEquality would turn every polynomial-space decision procedure into a polynomial-time one. Separation would establish that some problems admit modest memory requirements while resisting every deterministic polynomial-time algorithm. Status Open P versus PSPACE remains open. Recent space simulations and fixed-exponent time lower bounds do not separate polynomial time from polynomial space; conditional equality implications are not unconditional resolutions. Status reviewed Sources - Walter S. Brainerd and Lawrence H. Landweber et al., Computability, Decidability, Complexity, third edition. https://art.torvergata.it/bitstream/2108/32769/60437/ComputabilityDecidabilityComplexity.pdf Statement source - Simulating Time With Square-Root Space — Ryan Williams, February 24, 2025 https://eccc.weizmann.ac.il/report/2025/017/download Status source · Retrieved Sep 14, 2026 - Some Recent Developments in Space Complexity — R. Ryan Williams, MFCS 2026 https://drops.dagstuhl.de/entities/document/10.4230/LIPIcs.MFCS.2026.3 Status source · Retrieved Sep 14, 2026 - Some conditions implying if P=NP then P=PSPACE — Ismael Rodriguez February 10, 2026 https://arxiv.org/abs/2602.10073 Status source · Retrieved Sep 14, 2026 Problem familyClassical complexity separations · 5 problems in this edition. Family membership alone does not assert an implication. - 1 · P versus NP problem-p-versus-np - 13 · NP versus coNP Problem problem-np-versus-conp-problem - 14 · P versus PSPACE this problem - 34 · P versus NP ∩ coNP problem-top500-omission-w050054-p-versus-np-conp - 103 · Superpolynomial Lower Bounds for Frege Propositional Proof Systems problem-superpolynomial-lower-bounds-for-frege-propositional-proof-systems Ranking uncertainty90% source-fit bootstrap rank range: 11–15. Source-fit opponent count: 31. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 14. Rank 015 Tate Conjecture for Algebraic Cycles Algebraic subvarieties defined over the base field produce l-adic cohomology classes fixed by arithmetic symmetries after a normalization called a Tate twist. For smooth projective varieties over finitely generated fields, are all even-degree classes with this symmetry property generated by algebraic cycles over the same field, for every prime l different from the field’s characteristic?Geometry & topology Exact statementFor every smooth geometrically irreducible projective variety X over a field k finitely generated over its prime field, every codimension i, and every prime l different from char k , prove that the l-adic cycle-class map from codimension-i algebraic cycle classes defined over k, tensored with Q l, is surjective onto H^ 2i X bar,Q l i ^Gal k bar/k . Why it mattersA proof would recover every invariant class in the specified cohomology from algebraic cycles defined over the base field. A counterexample would identify an invariant class with no such algebraic description. Status Open · subcases solved The rational Tate conjecture remains open, with solved subcases. Integral counterexamples do not refute the stated rational cycle map. Kahn's unsuccessful surface approach does not prove the universal target; its HTML/history date discrepancy was not authenticated as a new revision. Status reviewed Sources - Li, What is the Tate conjecture?, map 2 and Conjecture 1 2013 . https://www.math.columbia.edu/~chaoli/docs/TateConjecture.html Statement source - An approach to the Tate conjecture for surfaces over a finite field — Bruno Kahn 2025 https://arxiv.org/abs/2505.07337 Status source · Retrieved Sep 14, 2026 - On the integral Tate conjecture for abelian varieties — J. S. Milne September 8, 2025 https://arxiv.org/abs/2509.06707 Status source · Retrieved Sep 14, 2026 - An approach to the Tate conjecture for surfaces over a finite field — Bruno Kahn, full text https://arxiv.org/html/2505.07337v1 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyTate Conjecture for Algebraic Cycles · 1 problem in this edition. Family membership alone does not assert an implication. - 15 · Tate Conjecture for Algebraic Cycles this problem Ranking uncertainty90% source-fit bootstrap rank range: 13–15. Source-fit opponent count: 33. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 15. Rank 016 Polynomial-time classical integer factorization Is there a deterministic classical algorithm that finds a nontrivial factor of every composite integer in time polynomial in the number of its binary digits?Theoretical computer science Exact statementDetermine whether there is a deterministic classical algorithm that, on input a composite integer N in binary, outputs a nontrivial factor of N in time polynomial in log N. Why it mattersAn algorithm would guarantee efficient factor finding for every composite input in this model. A negative result would establish that no deterministic classical algorithm can satisfy that polynomial-time bound. Status Open Deterministic classical polynomial-bit-time factorization remains open. Conditional improvements to exponents in the integer being factored are not polynomial time in its bit length, and quantum factoring is outside this target. Status reviewed Sources - Integer factoring and modular square roots https://www.sciencedirect.com/science/article/pii/S0022000015000768 Statement source - A number-theoretic conjecture implying faster algorithms for polynomial factorization and integer factorization — Chris Umans and Siki Wang 2025 https://arxiv.org/abs/2511.10851 Status source · Retrieved Sep 14, 2026 - Improved Separation Between Quantum and Classical Computers for Sampling and Functional Tasks — Marshall, Aaronson and Dunjko, CCC 2025 https://drops.dagstuhl.de/storage/00lipics/lipics-vol339-ccc2025/LIPIcs.CCC.2025.5/LIPIcs.CCC.2025.5.pdf Status source · Retrieved Sep 14, 2026 Problem familyPolynomial-time classical integer factorization · 1 problem in this edition. Family membership alone does not assert an implication. - 16 · Polynomial-time classical integer factorization this problem Ranking uncertainty90% source-fit bootstrap rank range: 16–18. Source-fit opponent count: 31. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 16. Rank 017 BQP Versus NP: Complete Language-Class Relation How do efficient quantum solving and efficient classical verification compare for yes-or-no problems? Does quantum computation with bounded error solve every problem with short, efficiently checkable evidence for yes answers—and do all problems it solves have such evidence?Theoretical computer science Exact statementDetermine the truth values of BOTH NP subseteq BQP and BQP subseteq NP for unrelativized, advice-free classes of total languages over {0,1}. NP consists of languages with polynomial-length binary witnesses and a deterministic polynomial-time verifier. BQP consists of languages decided by polynomial-time uniformly generated polynomial-size quantum circuits a deterministic Turing machine outputs Q n on 1^n , using Watrous's fixed Toffoli, Hadamard and phase-i gates with |0 ancillas and discarding, followed by a standard-basis output measurement. On every yes input acceptance is at least 2/3, and on every no input it is at most 1/3. No arbitrary real gate constants, advice or oracle are available. Establish equality, either proper containment, or incomparability. Why it mattersResolving both inclusions would determine whether these computational classes coincide, one strictly contains the other, or neither contains the other. Each outcome gives a different relationship between efficient verification and bounded-error quantum computation. Status Open Both unrelativized BQP/NP inclusions remain unresolved on the checked evidence. Random-permutation oracle separations do not settle the stated uniform total-language relationship. Status reviewed Sources - John Watrous, Quantum Computational Complexity, arXiv:0804.3401v1; language questions corroborated by Scott Aaronson lecture notes. https://arxiv.org/pdf/0804.3401v1 Statement source - Complexity Theory, Lecture 25: Quantum Computing 2 — Markus Krötzsch, January 27, 2025 https://iccl.inf.tu-dresden.de/w/images/4/4d/CT2024-Lecture-25-overlay.pdf Status source · Retrieved Sep 14, 2026 - Introduction to Quantum Information Science lecture notes — Scott Aaronson https://www.scottaaronson.com/qclec.pdf Status source · Retrieved Sep 14, 2026 - Random Permutations in Computational Complexity — Hitchcock, Sekoni and Shafei November 11, 2025 https://arxiv.org/abs/2511.08786 Status source · Retrieved Sep 14, 2026 Problem familyBQP Versus NP: Complete Language-Class Relation · 1 problem in this edition. Family membership alone does not assert an implication. - 17 · BQP Versus NP: Complete Language-Class Relation this problem Ranking uncertainty90% source-fit bootstrap rank range: 16–19. Source-fit opponent count: 37. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 17. Rank 018 Hardy-Littlewood Prime k-Tuple Conjecture Take a fixed finite pattern of integer offsets. If no prime divides the product of the shifted entries for every integer shift, do shifts making every entry prime occur with exactly the asymptotic frequency predicted by the Hardy–Littlewood formula, including its correction for divisibility?Number theory & arithmetic geometry Simultaneous prime values · 5 problems family-context-problem-hardy-littlewood-prime-k-tuple-conjecture Exact statementFor every finite admissible set H={h 1,...,h k} of distinct integers, prove that the number of n <= x for which every n+h i is prime is asymptotic to S H x/ log x ^k, where S H is the Hardy-Littlewood singular series, or disprove this formula. Why it mattersA proof would give quantitative frequencies for every admissible finite prime pattern. A counterexample would identify a pattern whose simultaneous-prime count departs from the proposed asymptotic law. Status Open The all-fixed-admissible-tuples Hardy–Littlewood asymptotic remains open. Published short-average progress does not establish every fixed tuple. The located bitstring-sieve full-proof claim lacks independent acceptance. Status reviewed Sources - k-Tuple Conjecture MathWorld https://mathworld.wolfram.com/k-TupleConjecture.html Statement source - Correlations of almost primes — Institut Henri Poincaré seminar abstract https://www.ihp.fr/en/node/3096 Status source · Retrieved Sep 14, 2026 - Proof of the Hardy-Littlewood K-tuple Conjecture in the Distribution of Numbers Coprime with the Primorial — Tim Samshuijzen, May 2026 revision https://rxiv.org/pdf/2501.0157v4.pdf Status source · Retrieved Sep 14, 2026 - Higher uniformity of arithmetic functions in short intervals II. Almost all intervals — Matomäki, Radziwiłł, Shao, Tao and Teräväinen 2026 https://link.springer.com/article/10.1007/s00222-026-01408-6 Status source · Retrieved Sep 14, 2026 Problem familySimultaneous prime values · 5 problems in this edition. Family membership alone does not assert an implication. - 18 · Hardy-Littlewood Prime k-Tuple Conjecture this problem - 20 · Twin Prime Conjecture problem-twin-prime-conjecture - 53 · Schinzel's Hypothesis H on Simultaneous Prime Values of Irreducible Polynomials problem-schinzel-s-hypothesis-h-on-simultaneous-prime-values-of-irreducible-polynomials - 58 · Dickson's Conjecture on Simultaneous Prime Values of Linear Forms problem-top500-omission-w045049-dickson-linear-prime-values - 81 · Bateman-Horn conjecture problem-bateman-horn-conjecture Ranking uncertainty90% source-fit bootstrap rank range: 16–20. Source-fit opponent count: 31. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 18. Rank 019 Unconditional Separation of Quantum and Classical Bounded-Error Polynomial Time BQP = BPP Can quantum computers solve some decision problem in polynomial time with bounded error that every classical randomized polynomial-time algorithm fails to solve with bounded error?Quantum information & computation Exact statementProve unconditionally that bounded-error quantum polynomial time BQP strictly contains bounded-error probabilistic polynomial time BPP . Why it mattersA proof would establish an unconditional separation between these quantum and classical computational resources. A negative resolution would equate their polynomial-time decision power when both are allowed bounded error. Status Open Unconditional unrelativized BQP versus BPP remains open. The September Fourier-hierarchy result is oracle-relative and does not establish an ordinary language-class separation. Status reviewed Sources - E. Bernstein, U. Vazirani, SIAM J. Comput. 26 5 :1411-1473, 1997. https://doi.org/10.1137/S0097539796300921 Statement source - Improved Separation Between Quantum and Classical Computers for Sampling and Functional Tasks — Marshall, Aaronson and Dunjko, CCC 2025 https://drops.dagstuhl.de/storage/00lipics/lipics-vol339-ccc2025/LIPIcs.CCC.2025.5/LIPIcs.CCC.2025.5.pdf Status source · Retrieved Sep 14, 2026 - Oracle Separations in the Fourier Hierarchy — Atul Mantri, September 10, 2026 https://arxiv.org/abs/2609.11830 Status source · Retrieved Sep 14, 2026 Problem familyUnconditional Separation of Quantum and Classical Bounded-Error Polynomial Time BQP = BPP · 1 problem in this edition. Family membership alone does not assert an implication. - 19 · Unconditional Separation of Quantum and Classical Bounded-Error Polynomial Time BQP = BPP this problem Ranking uncertainty90% source-fit bootstrap rank range: 17–21. Source-fit opponent count: 28. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 19. Rank 020 Twin Prime Conjecture Are there infinitely many pairs of prime numbers differing by exactly two, so that such pairs continue to occur however far along the number line we look?Number theory & arithmetic geometry Simultaneous prime values · 5 problems family-context-problem-twin-prime-conjecture Exact statementProve that there are infinitely many primes p such that p+2 is also prime. Why it mattersA proof would show that the exact gap of two recurs at arbitrarily large primes. It would turn this smallest positive gap between odd primes into an endlessly recurring feature of prime spacing. Status Open Twin primes remains open with uncorroborated 2026 proof-claim warnings. Repository hosting and proof titles are not independent mathematical acceptance; bounded gaps, finite computations and prime-pair analogues are not exact gap-two infinitude. Status reviewed Sources - Weisstein, E. W., Twin Prime Conjecture, From MathWorld - A Wolfram Resource, updated 16 August 2026. https://mathworld.wolfram.com/TwinPrimeConjecture.html Statement source - Elementary Number Theory — Henrik Bachmann, version 4, December 1, 2025 https://www.henrikbachmann.com/uploads/7/7/6/3/77634444/ent 2025 overviewnotes v4.pdf Status source · Retrieved Sep 14, 2026 - A New Look on the Twin-Prime Conjecture: The Proof of the Twin-Prime Conjecture — Woodward and Asheralieva https://researchportal.hw.ac.uk/en/publications/a-new-look-on-the-twin-prime-conjecture-the-proof-of-the-twin-pri/ Status source · Retrieved Sep 14, 2026 - Proof of the Hardy-Littlewood K-tuple Conjecture in the Distribution of Numbers Coprime with the Primorial — Tim Samshuijzen, May 2026 revision https://rxiv.org/pdf/2501.0157v4.pdf Status source · Retrieved Sep 14, 2026 - Twin Prime Conjecture 3rd Way — Taha Muhammad, June 19, 2026 https://www.cambridge.org/engage/coe/article-details/6a2d20d8d1922e37d572c5f8 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familySimultaneous prime values · 5 problems in this edition. Family membership alone does not assert an implication. Ranking uncertainty90% source-fit bootstrap rank range: 17–22. Source-fit opponent count: 31. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 20. Rank 021 Smooth four-dimensional Poincare conjecture A homotopy equivalence identifies spaces up to continuous deformation, without requiring a smooth match point by point. If a smooth, closed four-dimensional manifold has the homotopy type of the four-dimensional sphere, must there be a smooth bijection between them with a smooth inverse?Geometry & topology Smooth four-dimensional spheres · 2 problems family-context-problem-smooth-four-dimensional-poincare-conjecture Exact statementDetermine whether every smooth closed 4-manifold homotopy equivalent to the standard 4-sphere S^4 is diffeomorphic to S^4. Why it mattersA proof would show that the stated homotopy condition determines the sphere’s smooth structure. A counterexample would exhibit an exotic smooth four-sphere with a different diffeomorphism type. Status Open Smooth four-dimensional Poincare remains open. The reviewed X 41,189,73 project explicitly disclaims a peer-accepted counterexample and leaves geometric obligations outside its formal work. Restricted knot-route evidence does not decide the whole conjecture. Status reviewed Sources - Chris Gerig, No homotopy 4-sphere invariants using ECH = SWF, Algebraic & Geometric Topology 21 2021 , Conjecture 1.1. https://arxiv.org/pdf/1905.10938 Statement source - smooth4pc-t73-lean: Independent Review https://github.com/toffee-desuwa/smooth4pc-t73-lean/blob/main/docs/INDEPENDENT REVIEW.md Status source · Retrieved Sep 14, 2026 - The Smooth Four-Dimensional Poincare Conjecture https://epoch.ai/frontiermath/open-problems/smooth-poincare-4d Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familySmooth four-dimensional spheres · 2 problems in this edition. Family membership alone does not assert an implication. - 21 · Smooth four-dimensional Poincare conjecture this problem - 165 · Smooth Four-Dimensional Schoenflies Problem problem-smooth-four-dimensional-schoenflies-problem Ranking uncertainty90% source-fit bootstrap rank range: 20–32. Source-fit opponent count: 34. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 21. Rank 022 Hilbert's Tenth Problem over the Rationals Is there an algorithm that always halts and correctly decides whether any given integer-coefficient polynomial, in any finite number of variables, has a rational solution?Logic, foundations & set theory Hilbert’s tenth problem · 2 problems family-context-problem-hilberts-tenth-problem-over-the-rationals Exact statementThere exists a total algorithm which, given any finite encoding of a multivariate polynomial f in Z x 1,...,x n with arbitrary finite n and integer coefficients, halts and correctly decides whether there is q in Q^n with f q =0. Standard computably interconvertible polynomial encodings are equivalent; no uniform running-time complexity bound is part of the assertion. Why it mattersA positive answer would supply a universal decision procedure for rational solvability, with no efficiency guarantee. A negative answer would establish that no algorithm can settle every polynomial input while always terminating. Status Open Hilbert's tenth problem over Q remains open. Results for finitely generated rings or languages with extra height predicates do not give a decision/undecidability result for ordinary rational Diophantine solvability. Status reviewed Sources - Sylvy Anscombe, Valentijn Karemaker, Zeynep Kisakürek, Vlerë Mehmeti, Margherita Pagano and Laura Paladino, A survey of local-global methods for Hilbert's Tenth Problem, abstract and Section1. https://arxiv.org/pdf/2309.14987 Statement source - Hilbert's tenth problem for finitely generated rings https://arxiv.org/html/2602.04468v1 Status source · Retrieved Sep 14, 2026 - Effectivity for existence of rational points is undecidable https://arxiv.org/abs/2311.01958 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyHilbert’s tenth problem · 2 problems in this edition. Family membership alone does not assert an implication. - 22 · Hilbert's Tenth Problem over the Rationals this problem - 430 · Hilbert's Tenth Problem over C t with named t problem-provisional-top500-omission-waves-010-014.018 Ranking uncertainty90% source-fit bootstrap rank range: 20–30. Source-fit opponent count: 42. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 22. Rank 023 The NC versus P Problem for Efficient Parallel Computation Can every decision problem solvable in polynomial time be solved with polynomially many parallel processors in only a power of log n sequential steps, where n is the input length? The parallel computations must come from a uniform algorithmic construction.Theoretical computer science Exact statementDetermine whether NC equals P, where NC consists of decision problems solvable by uniform polynomial-size circuit families of polylogarithmic depth and P consists of decision problems solvable in deterministic polynomial time. Why it mattersEquality would allow every polynomial-time decision problem to be organized into shallow uniform circuits. Separation would identify a problem whose polynomial-time computation cannot be parallelized to that depth within polynomial size. Status Open NC versus P remains open with a dated exact-status-source warning. Recent cellular-automaton or restricted-model classifications do not separate the same ordinary language classes. Status reviewed Sources - J. Heemstra, J. Martens, and A. Wijs, Evaluating Massively Parallel Algorithms for DFA Minimisation, Equivalence Checking and Inclusion Checking, Introduction 2026 . https://arxiv.org/abs/2508.20735 Statement source - Parallel Computation, Matt Williamson https://community.wvu.edu/~krsubramani/courses/sp09/cc/lecnotes/parallel.pdf Status source · Retrieved Sep 14, 2026 - Complexity of the Freezing Majority Rule with L-shaped Neighborhoods https://arxiv.org/html/2509.16065v1 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyThe NC versus P Problem for Efficient Parallel Computation · 1 problem in this edition. Family membership alone does not assert an implication. - 23 · The NC versus P Problem for Efficient Parallel Computation this problem Ranking uncertainty90% source-fit bootstrap rank range: 21–30. Source-fit opponent count: 49. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 23. Rank 024 Grothendiecks Generalized Hodge Conjecture Hodge coniveau measures how far Hodge theory suggests cohomology classes can be concentrated on a smaller subset. On a smooth complex projective variety, must every rational Hodge substructure of coniveau at least c be supported on an algebraic subset of codimension at least c?Number theory & arithmetic geometry Hodge conjectures · 3 problems family-context-problem-grothendiecks-generalized-hodge-conjecture Exact statementFor every smooth projective complex variety X and integers k,c, prove that the largest rational Hodge substructure of H^k X,Q with Hodge coniveau at least c equals the subspace of classes that vanish on X minus some closed algebraic subset of codimension at least c. Why it mattersA proof would identify a condition on Hodge structures with a geometric support condition in every specified degree and codimension. A counterexample would show that the Hodge condition can predict support that no suitable algebraic subset provides. Status Open · subcases solved Generalized Hodge remains open, with solved subcases. July 2026 treatment retains an unresolved general conjecture; moduli-space and specified coniveau cases do not cover every smooth projective variety. Status reviewed Sources - Claire Voisin, Hodge and generalized Hodge conjectures, coniveau and algebraic cycles, Journal of Open Mathematical Problems 1 1 2025 , 16-51. https://jomprob.org/index.php/jomp/article/view/Vol-1Issue-1Paper-2 Statement source - Incidence equivalence, a survey https://arxiv.org/abs/2607.22233 Status source · Retrieved Sep 14, 2026 - On the Hodge and Tate conjectures for moduli spaces of curves https://arxiv.org/html/2605.20453v1 Status source · Retrieved Sep 14, 2026 Problem familyHodge conjectures · 3 problems in this edition. Family membership alone does not assert an implication. - 4 · Hodge Conjecture problem-hodge-conjecture - 24 · Grothendiecks Generalized Hodge Conjecture this problem - 90 · Deligne's Absolute Hodge Conjecture problem-top500-omission-w045049-absolute-hodge-conjecture Ranking uncertainty90% source-fit bootstrap rank range: 21–33. Source-fit opponent count: 32. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 24. Rank 025 Binary Goldbach Problem for Even Integers Can every even integer greater than two be written as the sum of two prime numbers, allowing the same prime to be used twice?Number theory & arithmetic geometry Exact statementProve or disprove that every even integer n greater than 2 is the sum of two primes. Why it mattersA proof would guarantee that two prime summands suffice to represent every even integer above two. It would give a universal additive representation across all those integers, however large. Status Open Goldbach remains open. July/August 2026 sources distinguish the conjecture from almost-prime results; the revised uniformity preprint explicitly does not yield pointwise Goldbach. Status reviewed Sources - C. K. Caldwell, "Prime Conjectures and Open Questions," The PrimePages, https://t5k.org/notes/conjectures/. https://t5k.org/notes/conjectures/ Statement source - The exceptional set of the Goldbach problem https://arxiv.org/html/2607.27282v2 Status source · Retrieved Sep 14, 2026 - Theorem 1+1.9 on the Goldbach Conjecture https://arxiv.org/abs/2606.05224 Status source · Retrieved Sep 14, 2026 - Restricted Goldbach Sums in Arithmetic Progressions: Local Obstructions, an Explicit Almost-All Framework, and a General-Modulus Extension https://www.preprints.org/manuscript/202607.2077 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyBinary Goldbach Problem for Even Integers · 1 problem in this edition. Family membership alone does not assert an implication. - 25 · Binary Goldbach Problem for Even Integers this problem Ranking uncertainty90% source-fit bootstrap rank range: 21–37. Source-fit opponent count: 29. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 25. Rank 026 P versus BPP Does BPP = P? Can every decision problem solved by a randomized polynomial-time algorithm with bounded error on both answers also be solved deterministically in polynomial time?Cryptography, coding, information & optimization Exact statementDetermine whether every language computable by a probabilistic polynomial-time algorithm with bounded two-sided error BPP can be computed by a deterministic polynomial-time algorithm, i.e., whether BPP = P. Why it mattersEquality would remove randomness from every computation in this class while preserving polynomial-time solvability. Separation would identify a decision problem for which bounded-error randomization provides power beyond deterministic polynomial time. Status Open P versus BPP remains open. Recent pseudorandom-generator and derandomization advances retain additional assumptions, not a deterministic simulation of every BPP language. Status reviewed Sources - S. Vadhan, 'Pseudorandomness', Foundations and Trends in Theoretical Computer Science 7 1-3 :1-336, 2012. https://people.seas.harvard.edu/~salil/pseudorandomness/pseudorandomness-Aug12.pdf Statement source - CS 6810: Theory of Computing, Lecture 15, March 17, 2026 https://www.cs.cornell.edu/courses/cs6810/2026sp/lec15.pdf Status source · Retrieved Sep 14, 2026 - Pseudorandomness Beating the Hybrid Argument for Insensitive Algorithms https://eccc.weizmann.ac.il/report/2026/082/ Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyP versus BPP Does BPP = P? · 1 problem in this edition. Family membership alone does not assert an implication. - 26 · P versus BPP Does BPP = P? this problem Ranking uncertainty90% source-fit bootstrap rank range: 22–36. Source-fit opponent count: 29. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 26. Rank 027 Schanuel's conjecture If n complex numbers have no nonzero rational linear relation, must at least n of those numbers and their exponentials be algebraically independent, meaning that no nonzero polynomial with rational coefficients relates them?Number theory & arithmetic geometry Exponential transcendence · 3 problems family-context-problem-schanuel-conjecture Exact statementProve or disprove that if z 1,...,z n in C are linearly independent over Q, then the field Q z 1,...,z n,e^{z 1},...,e^{z n} has transcendence degree at least n over Q. Why it mattersA proof would force at least n algebraically independent quantities into every field generated this way. A counterexample would reveal an exponential tuple with more algebraic dependence than the bound allows. Status Open Schanuel remains open. Reformulations and almost-everywhere Schanuel properties do not establish its universal algebraic-independence conclusion. Status reviewed Sources - Schanuel's Conjecture and the Transcendence of Power Towers https://www.mdpi.com/2227-7390/9/7/717 Statement source - Combining the conjectures of Schanuel and Zilber-Pink https://content.ems.press/assets/public/full-texts/serials/rlm/36/3/14299588/online/10.4171-rlm-1086.pdf Status source · Retrieved Sep 14, 2026 - Schanuel Property for Elliptic and Quasi-Elliptic Functions https://webusers.imj-prg.fr/~michel.waldschmidt/articles/pdf/SchanuelPropertyAlmostAll.pdf Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyExponential transcendence · 3 problems in this edition. Family membership alone does not assert an implication. - 27 · Schanuel's conjecture this problem - 176 · Algebraic Independence of e and pi problem-algebraic-independence-of-e-and-pi - 379 · Four Exponentials Conjecture problem-four-exponentials-conjecture Ranking uncertainty90% source-fit bootstrap rank range: 21–38. Source-fit opponent count: 30. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 27. Rank 028 Matrix-Multiplication Exponent Equals Two Can two n-by-n rational matrices be multiplied exactly, without division, with a number of arithmetic operations arbitrarily close to quadratic: O n^ 2 + ε for every positive ε? Each rational arithmetic operation counts as one step, regardless of the number of bits in its operands.Theoretical computer science Exact statementLet R Q n be the tensor rank over Q of the bilinear map multiplying two n by n rational matrices: the least r for which AB is a sum of r products alpha j A beta j B C j, with rational linear forms alpha j, beta j and rational output matrices C j. Set omega Q = inf{tau in R : R Q n = O n^tau }. Prove omega Q = 2. Equivalently, for every epsilon 0, exact multiplication admits division-free arithmetic circuits over Q of size O n^ 2+epsilon , counting rational additions, subtractions, multiplications and constant scalar operations at unit cost. Circuits need not be uniformly generated and rational coefficient bit lengths are not charged. Constants and thresholds may depend on epsilon. Literal O n^2 attainment is not required. Why it mattersA proof would allow exact matrix multiplication with arithmetic exponents arbitrarily close to two and carry through applicable algebraic reductions to other computations. The statement measures arithmetic operations; it leaves coefficient bit costs and literal quadratic attainment outside its guarantee. Status Open The exact rational-field matrix multiplication exponent target remains open. The August 2026 claimed exponent below 2.371177 is not exponent 2; complex-field results require field/scope review before transfer to Q. Status reviewed Sources - Markus Bläser, Fast Matrix Multiplication, Theory of Computing Graduate Surveys 5 2013 . https://theoryofcomputing.org/articles/gs005/gs005.pdf Statement source - Improving the matrix multiplication exponent with modern optimization and AlphaEvolve https://arxiv.org/abs/2608.16884 Status source · Retrieved Sep 14, 2026 - Finite Matrix Multiplication Algorithms from Infinite Groups https://drops.dagstuhl.de/storage/00lipics/lipics-vol325-itcs2025/html/LIPIcs.ITCS.2025.18/LIPIcs.ITCS.2025.18.html Status source · Retrieved Sep 14, 2026 Problem familyMatrix-Multiplication Exponent Equals Two · 1 problem in this edition. Family membership alone does not assert an implication. - 28 · Matrix-Multiplication Exponent Equals Two this problem Ranking uncertainty90% source-fit bootstrap rank range: 22–35. Source-fit opponent count: 46. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 28. Rank 029 Unique Games Conjecture In a graph-labeling game, each edge requires the two labels to match through a one-to-one rule. Is distinguishing games where almost all rules can be satisfied from those where only a tiny fraction can be satisfied NP-hard, however small these two error thresholds are, with a fixed label set chosen for each pair of thresholds?Theoretical computer science Unique Games and approximation thresholds · 2 problems family-context-problem-khots-unique-games-conjecture Exact statementFor every fixed pair of real constants epsilon, delta 0 with epsilon + delta < 1, there exists an integer k = 2, depending only on epsilon and delta, such that the following promise problem is NP-hard under deterministic polynomial-time many-one reductions. An instance is a finite directed constraint graph G= V,E with E nonempty and, for each edge e= v,w , a permutation pi e of k . For a labeling L:V- k , an edge is satisfied if pi e L v =L w ; val G is the largest fraction of satisfied edges over all labelings. Distinguish val G = 1-epsilon from val G <= delta, with no requirement on instances between these thresholds. The graph is the growing input; epsilon, delta and k are fixed constants. NP-hardness means that every language in NP has such a reduction mapping yes inputs to the first promised set and no inputs to the second. Why it mattersA proof would establish this hardness gap and support approximation thresholds obtained through Unique Games reductions. Claims of efficient-algorithm impossibility from NP-hardness would still require the assumption P ≠ NP. Status Open Classical UGC remains open. The accessible Hallelujah preprint's graphwise ratio strictly below 2 does not give the fixed uniform 2-epsilon improvement needed for its claimed contradiction. The linked journal full text was inaccessible; independent acceptance of a full disproof was not established. Status reviewed Sources - Subhash Khot, On the Unique Games Conjecture, author-hosted survey. https://cs.nyu.edu/~khot/papers/UGCSurvey.pdf Statement source - Towards a Proof of the 2-to-1 Games Conjecture https://theoryofcomputing.org/articles/v021a011/v021a011.pdf Status source · Retrieved Sep 14, 2026 - An Approximate Solution to the Minimum Vertex Cover Problem: The Hallelujah Algorithm https://www.preprints.org/manuscript/202510.2392 Status source · Retrieved Sep 14, 2026 - Vertex Cover Might be Hard to Approximate to within 2-epsilon https://cims.nyu.edu/~regev/papers/vc hard.pdf Status source · Retrieved Sep 14, 2026 - Sharp Hardness for MAX-3-CUT and Quantum MAX-CUT https://arxiv.org/abs/2608.00333 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyUnique Games and approximation thresholds · 2 problems in this edition. Family membership alone does not assert an implication. - 29 · Unique Games Conjecture this problem - 233 · Unconditional 2−epsilon Inapproximability of Minimum Vertex Cover problem-unconditional-gap-two-vertex-cover-hardness Ranking uncertainty90% source-fit bootstrap rank range: 22–37. Source-fit opponent count: 49. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 29. Rank 030 Generalized Ramanujan Conjecture for GL n Automorphic representations connect arithmetic with highly symmetric functions. For every cuspidal representation of GL n over a number field with unitary central character, must its component at every place satisfy the sharp local condition called temperedness?Number theory & arithmetic geometry Exact statementProve that for every number field F and every cuspidal automorphic representation pi of GL n A F with unitary central character, each local component pi v is tempered at every place v. Why it mattersA proof would force the global cuspidal and unitary hypotheses to constrain local behavior at every place. It would rule out non-tempered local components throughout the specified automorphic family. Status Open · subcases solved Generalized Ramanujan remains open, with solved subcases. Recent analytic consequences assume the conjecture and restricted automorphic cases do not prove all GL n targets; an inaccessible purported full proof was not accepted. Status reviewed Sources - Peter Sarnak, Notes on the Generalized Ramanujan Conjectures, in Harmonic Analysis, the Trace Formula, and Shimura Varieties. https://web.math.princeton.edu/~sarnak/Preprints/SarnakFieldsNotes.pdf Statement source - A note on Sarnak's density hypothesis for Sp4 https://ems.press/content/serial-article-files/53056 Status source · Retrieved Sep 14, 2026 - Regularized spectral expansion of a Rankin-Selberg period and moments https://arxiv.org/abs/2609.01358 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyGeneralized Ramanujan Conjecture for GL n · 1 problem in this edition. Family membership alone does not assert an implication. - 30 · Generalized Ramanujan Conjecture for GL n this problem Ranking uncertainty90% source-fit bootstrap rank range: 23–36. Source-fit opponent count: 42. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 30. Rank 031 Anderson Model Extended States in Dimension at Least Three In a lattice of dimension at least three, can a quantum particle in a weak random potential have a conducting regime, separated from localized behavior by a mobility edge? The mathematical target is an absolutely continuous part of the Anderson model’s energy spectrum.Mathematical physics Exact statementProve that the Anderson model, the discrete random Schroedinger operator on Z^d with d = 3, has a regime of absolutely continuous spectrum, equivalently that a conducting phase and a mobility edge exist at weak disorder, so that the metal-insulator transition is established. Why it mattersA proof would establish the specified spectral regime and the associated transition sought in this random-operator model. It would connect the model’s weak-disorder behavior to the conducting phase described in the target. Status Open The lattice Anderson extended-states target remains open, explicitly so in a September 7 primary paper. Bethe-lattice theorems and finite-timescale lattice diffusion do not establish absolutely continuous spectrum on Z^d at fixed positive disorder. The claimed equivalence with conduction/mobility edges remains unverified. Status reviewed Sources - W. Kirsch, "An Invitation to Random Schrodinger Operators", in Random Schrodinger Operators, Panoramas et Syntheses 25, Societe Mathematique de France, 2008; arXiv:0709.3707. https://arxiv.org/abs/0709.3707 Statement source - A Note on Extended States on the Bethe Lattice https://arxiv.org/html/2609.06987v1 Status source · Retrieved Sep 14, 2026 - Self-consistent equations and quantum diffusion for the Anderson model https://arxiv.org/abs/2506.06468 Status source · Retrieved Sep 14, 2026 Problem familyAnderson Model Extended States in Dimension at Least Three · 1 problem in this edition. Family membership alone does not assert an implication. - 31 · Anderson Model Extended States in Dimension at Least Three this problem Ranking uncertainty90% source-fit bootstrap rank range: 23–36. Source-fit opponent count: 53. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 31. Rank 032 Valiants Algebraic P versus NP Conjecture VP = VNP The permanent uses the determinant’s sum over permutations but gives every term a plus sign. Over fields of characteristic zero, must arithmetic circuits computing n-by-n permanents exactly require more than polynomially many operations as n grows?Theoretical computer science Algebraic computation lower bounds · 3 problems family-context-problem-valiants-algebraic-p-versus-np-conjecture-vp-vnp Exact statementProve that the permanent polynomial family cannot be computed by polynomial-size algebraic circuits over fields of characteristic zero, equivalently that VP is not equal to VNP. Why it mattersA proof would separate the algebraic complexity classes VP and VNP through a circuit lower bound for the permanent. Polynomial-size circuits for the permanent would refute that separation. Status Open VP versus VNP remains open over the stated characteristic-zero model. January 2026 primary work still poses the unrestricted question; symmetry-restricted circuits and unconstructed elusive-function reductions do not resolve it. Status reviewed Sources - Dwivedi, Pago, and Seppelt, Lower Bounds in Algebraic Complexity via Symmetry and Homomorphism Polynomials, STOC 2026, Introduction. https://www.prateekdwivedi.in/papers/symACT-confversion.pdf Statement source - Symmetric Algebraic Circuits and Homomorphism Polynomials https://drops.dagstuhl.de/storage/00lipics/lipics-vol362-itcs2026/LIPIcs.ITCS.2026.46/LIPIcs.ITCS.2026.46.pdf Status source · Retrieved Sep 14, 2026 - Arithmetic circuit lower bounds from sumset expansion https://eccc.weizmann.ac.il/report/2026/138/ Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyAlgebraic computation lower bounds · 3 problems in this edition. Family membership alone does not assert an implication. - 32 · Valiants Algebraic P versus NP Conjecture VP = VNP this problem - 62 · Permanent versus Determinant over C problem-permanent-versus-determinant-conjecture - 340 · Koiran's Real Tau-Conjecture on Real Roots of Sums of Products of Sparse Polynomials problem-koiran-real-tau-conjecture Ranking uncertainty90% source-fit bootstrap rank range: 26–36. Source-fit opponent count: 48. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 32. Rank 033 Basing One-Way Functions on NP-Hardness Can worst-case NP-hardness provide a rigorous foundation for one-way functions—functions that are easy to compute but hard to reverse on random inputs—through a reduction that makes P ≠ NP imply their existence?Cryptography, coding, information & optimization Exact statementDetermine whether one-way functions can be based on NP-hardness, i.e., whether there is a reduction from a worst-case NP-complete or NP-hard decision problem to the task of average-case inverting a polynomial-time computable function, so that P = NP would imply the existence of one-way functions. Why it mattersSuch a reduction would derive average-case inversion hardness from the stated worst-case complexity separation. It would connect the existence of one-way functions to the assumption P ≠ NP through a precise computational implication. Status Open Constructing one-way functions from worst-case NP hardness alone remains open. Located meta-complexity, zero-knowledge and derandomization characterizations add unproved hypotheses beyond P = NP. Status reviewed Sources - A. Akavia, O. Goldreich, S. Goldwasser and D. Moshkovitz, 'On basing one-way functions on NP-hardness', Proc. 38th Annual ACM Symp. on Theory of Computing STOC , 2006, pp. 701-710. DOI:10.1145/1132516.1132614. https://api.openalex.org/works/doi:10.1145/1132516.1132614 Statement source - Non-Trivial Zero-Knowledge Implies One-Way Functions https://arxiv.org/abs/2602.17651 Status source · Retrieved Sep 14, 2026 - Hardness Along the Boundary: Towards One-Way Functions from the Worst-case Hardness of Time-Bounded Kolmogorov Complexity https://eccc.weizmann.ac.il/report/2025/125/ Status source · Retrieved Sep 14, 2026 - Capturing One-Way Functions via NP-Hardness of Meta-Complexity https://eccc.weizmann.ac.il/report/2023/037/ Status source · Retrieved Sep 14, 2026 Problem familyBasing One-Way Functions on NP-Hardness · 1 problem in this edition. Family membership alone does not assert an implication. - 33 · Basing One-Way Functions on NP-Hardness this problem Ranking uncertainty90% source-fit bootstrap rank range: 26–39. Source-fit opponent count: 43. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 33. Rank 034 P versus NP ∩ coNP If both yes and no answers to a decision problem have short evidence that can be checked efficiently, must a deterministic computer be able to solve the problem in polynomial time?Theoretical computer science Classical complexity separations · 5 problems family-context-problem-top500-omission-w050054-p-versus-np-conp Exact statementDetermine whether P = NP ∩ coNP for ordinary uniform deterministic and nondeterministic polynomial-time Turing-machine classes of total languages over a finite alphabet; equivalently, whether every language with polynomially checkable certificates for both membership and nonmembership has a deterministic polynomial-time decision algorithm. Why it mattersEquality would turn efficient certification on both sides into efficient decision-making for every such language. A separation would exhibit a language whose two kinds of certificates are easy to check but whose answers are not deterministically computable in polynomial time. Status Open The ordinary uniform P = NP intersect coNP question remains open on the checked evidence. The November 2025 random-permutation separation is relativized and does not settle the ordinary question. Status reviewed Sources - Scott Aaronson, PHYS771 Lecture 6: P, NP, and Friends https://www.scottaaronson.com/democritus/lec6.html Statement source - Very few problems are in NP intersect coNP but not known to be in P. What to make of that? https://blog.computationalcomplexity.org/2024/09/very-few-problems-are-in-np-intersect.html?m=0 Status source · Retrieved Sep 14, 2026 - Random Permutations in Computational Complexity https://arxiv.org/abs/2511.08786 Status source · Retrieved Sep 14, 2026 Problem familyClassical complexity separations · 5 problems in this edition. Family membership alone does not assert an implication. - 1 · P versus NP problem-p-versus-np - 13 · NP versus coNP Problem problem-np-versus-conp-problem - 14 · P versus PSPACE problem-p-versus-pspace - 34 · P versus NP ∩ coNP this problem - 103 · Superpolynomial Lower Bounds for Frege Propositional Proof Systems problem-superpolynomial-lower-bounds-for-frege-propositional-proof-systems Ranking uncertainty90% source-fit bootstrap rank range: 24–45. Source-fit opponent count: 37. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 34. Rank 035 Unconditional Separation of BQP from the Polynomial Hierarchy Can an efficient quantum computer solve a yes-or-no problem beyond every fixed level of the polynomial hierarchy, which allows successively nested layers of existential and universal verification? The quantum algorithm must have bounded error and use no oracle.Quantum information & computation Exact statementProve in the standard unrelativized model that bounded-error quantum polynomial time BQP is not contained in the polynomial hierarchy PH. Why it mattersA proof would place some efficient quantum computation beyond the entire polynomial hierarchy. The opposite conclusion would contain all bounded-error quantum polynomial-time decision problems within that hierarchy. Status Open Unrelativized BQP versus PH remains open with a dated exact-status-source warning. Recent oracle, query and promise-language results do not exhibit the required ordinary total language outside PH. Status reviewed Sources - Scott Aaronson, BQP and the Polynomial Hierarchy, arXiv:0910.4698, Abstract 2009 . https://arxiv.org/abs/0910.4698 Statement source - BQP and the Polynomial Hierarchy https://www.scottaaronson.com/papers/stoc114-aaronson.pdf Status source · Retrieved Sep 14, 2026 - IQP circuits for 2-Forrelation https://arxiv.org/abs/2604.15248 Status source · Retrieved Sep 14, 2026 - Promises should be taken seriously: On relativization with promise problems https://arxiv.org/abs/2609.07945 Status source · Retrieved Sep 14, 2026 Problem familyUnconditional Separation of BQP from the Polynomial Hierarchy · 1 problem in this edition. Family membership alone does not assert an implication. - 35 · Unconditional Separation of BQP from the Polynomial Hierarchy this problem Ranking uncertainty90% source-fit bootstrap rank range: 25–47. Source-fit opponent count: 44. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 35. Rank 036 Extension of Kroneckers Theorem on Abelian Fields to Any Algebraic Base Field Can all finite abelian extensions of any number field be generated by values of explicit analytic or arithmetic functions? These are extensions whose field symmetries commute; the goal is an explicit recipe for their generators.Number theory & arithmetic geometry Explicit class field theory and Stark conjectures · 3 problems family-context-problem-extension-of-kroneckers-theorem-on-abelian-fields-to-any-algebraic-base-field Exact statementFor each number field K, construct generators for its finite abelian extensions as special values of explicitly defined analytic or arithmetic functions associated with K, extending roots of unity over Q and complex multiplication over imaginary quadratic fields. Why it mattersA construction would give explicit generators for every extension in scope through a common special-value description. An obstruction for one number field would show a limit to that proposed universal construction. Status Open · subcases solved Hilbert's twelfth problem over arbitrary number fields remains open, with solved subcases. Dasgupta–Kakde and the independent ICMS account cover a totally real p-adic construction, not every number field or the original complex-analytic formulation. Status reviewed Sources - International Centre for Mathematical Sciences, Recent progress on Hilbert's 12th problem 2024 https://icms.ac.uk/archive/workshop/recent-progress-on-hilberts-12th-problem/ Statement source - Brumer–Stark Units and Explicit Class Field Theory https://services.math.duke.edu/~dasgupta/papers/Integral-Gross-Stark.pdf Status source · Retrieved Sep 14, 2026 - Recent Progress on Hilbert's 12th Problem https://icms.ac.uk/archive/workshop/recent-progress-on-hilberts-12th-problem/ Status source · Retrieved Sep 14, 2026 Problem familyExplicit class field theory and Stark conjectures · 3 problems in this edition. Family membership alone does not assert an implication. - 36 · Extension of Kroneckers Theorem on Abelian Fields to Any Algebraic Base Field this problem - 160 · Tate's rational leading-term Stark conjecture problem-tates-rational-leading-term-stark-conjecture - 197 · Rank One Abelian Stark Conjecture problem-rank-one-abelian-stark-conjecture Ranking uncertainty90% source-fit bootstrap rank range: 31–44. Source-fit opponent count: 51. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 36. Rank 037 Prove the Penrose inequality or present a counterexample For isolated gravitational initial data in three spatial dimensions obeying the dominant energy condition, must an outermost apparent horizon force the total mass to satisfy m ≥ √ A/ 16π , where A is the least area enclosing it, with equality characterized by Schwarzschild geometry?Mathematical physics Exact statementLet M,g,k be a three-dimensional asymptotically flat initial data set for the Einstein equations satisfying the dominant energy condition and containing an outermost apparent horizon. If A is the least area needed to enclose that horizon and m ADM is the ADM mass, prove the sharp spacetime Penrose inequality m ADM = sqrt A/ 16 pi , with the appropriate Schwarzschild rigidity statement, or construct initial data satisfying the hypotheses that violate the inequality. Why it mattersA proof would give a sharp lower bound on total mass from the least area enclosing the horizon and identify the equality geometry. A counterexample would show that the stated geometric and energy conditions permit a mass below the proposed area-based bound. Status Open · subcases solved The sharp spacetime Penrose inequality remains open, with solved subcases. Located MOTS and geometric extensions retain extra hypotheses or a weaker constant; no full sharp unconditional proof was verified. Status reviewed Sources - D. Xu, 'The Spacetime Penrose Inequality: Conditional Results for Stable MOTS and General Trapped Surfaces,' arXiv:2512.04137v5 2025 . https://arxiv.org/abs/2512.04137 Statement source - The Spacetime Penrose Inequality: Conditional Results for Stable MOTS and General Trapped Surfaces https://arxiv.org/abs/2512.04137 Status source · Retrieved Sep 14, 2026 - The spacetime Penrose inequality under a quasi final state hypothesis https://arxiv.org/abs/2605.18730 Status source · Retrieved Sep 14, 2026 - Proof of the Spacetime Penrose Inequality With Suboptimal Constant in the Asymptotically Flat and Asymptotically Hyperboloidal Regimes https://arxiv.org/abs/2504.10641 Status source · Retrieved Sep 14, 2026 Problem familyProve the Penrose inequality or present a counterexample · 1 problem in this edition. Family membership alone does not assert an implication. - 37 · Prove the Penrose inequality or present a counterexample this problem Ranking uncertainty90% source-fit bootstrap rank range: 31–46. Source-fit opponent count: 40. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 37. Rank 038 Higher-dimensional Euclidean Kakeya conjecture In four or more dimensions, must a compact set containing a unit line segment in every direction have full Hausdorff dimension? This asks for full fractal dimension, even when the set need not have positive volume.Analysis & PDE Kakeya, restriction and smoothing estimates · 4 problems family-context-problem-higher-dimensional-euclidean-kakeya-conjecture Exact statementProve or disprove that for every integer n = 4, every compact set E in R^n that contains a unit line segment in every direction has Hausdorff dimension n. Why it mattersA proof would show that accommodating all directions forces full Hausdorff dimension in these spaces. A counterexample would contain every required segment within a compact set of smaller dimension. Status Open · subcases solved The stated higher-dimensional Euclidean Kakeya target n = 4 remains open, with solved subcases. The solved three-dimensional theorem is outside that range. A February positive-volume preprint is unaccepted and does not overcome the arbitrary zero-measure Kakeya-set issue; September lifting results remain conditional. Status reviewed Sources - Mukul Rai Choudhuri, Revista Matematica Iberoamericana 42 2026 , 1227-1256 https://ems.press/journals/rmi/articles/14299708 Statement source - Fourier analytic properties of Kakeya sets in finite fields https://londmathsoc.onlinelibrary.wiley.com/doi/10.1112/blms.70367 Status source · Retrieved Sep 14, 2026 - Lifting curved Kakeya sets to linear Kakeya sets https://arxiv.org/abs/2609.06299 Status source · Retrieved Sep 14, 2026 - A Unified Geometric Algebra Framework for the Kakeya Conjecture in All Dimensions and Links to the Riemann Zeta Function https://www.preprints.org/manuscript/202602.1025 Status source · Retrieved Sep 14, 2026 Problem familyKakeya, restriction and smoothing estimates · 4 problems in this edition. Family membership alone does not assert an implication. - 38 · Higher-dimensional Euclidean Kakeya conjecture this problem - 45 · Fourier restriction conjecture problem-fourier-restriction-conjecture - 202 · Bochner-Riesz conjecture problem-bochner-riesz-conjecture - 206 · Local Smoothing Conjecture for Wave Equations problem-local-smoothing-conjecture-for-wave-equations Ranking uncertainty90% source-fit bootstrap rank range: 33–46. Source-fit opponent count: 53. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 38. Rank 039 Modularity of Elliptic Curves over Number Fields End K E =Z For any number field K, does every elliptic curve whose K-defined endomorphisms are just integer multiplications have the predicted automorphic realization: an embedding of its finite-place representation into the cohomology of the associated arithmetic spaces?Number theory & arithmetic geometry Langlands functoriality and modularity · 2 problems family-context-problem-modularity-of-elliptic-curves-over-number-fields Exact statementFor every number field K and every elliptic curve E/K with End K E =Z, prove that pi E admits a GL 2 A K^infinity -equivariant injection into A tensor Q C, exactly as in Thorne's Conjecture 2.6. Here A is the direct limit over open compact U of H^ Y U,Q for the adelic locally symmetric spaces defined on page 647, and pi E is the restricted tensor product at all finite places v of rec {K v}^{-1} r v tensor |.|^{1/2},N v , with r v,N v attached to E in the source's geometric-Frobenius normalization on pages 646 and 649. Why it mattersA proof would realize the arithmetic data encoded by the curve’s specified finite-place representation inside the cohomology of those spaces. The injection would respect the adelic group action, connecting the curve’s local representations with this geometric construction. Status Open · subcases solved Modularity over arbitrary number fields remains open, with solved subcases for the stated End K E =Z scope. Totally real tower and odd-degree uniformization results are restricted or conditional; End K E =Z is not silently replaced by geometrically non-CM. Status reviewed Sources - J. A. Thorne, Elliptic curves and modularity, 8ECM proceedings, EMS Press, 2023, DOI 10.4171/8ECM/12 https://ems.press/content/book-chapter-files/28276 Statement source - Elliptic curves and modularity https://ems.press/content/book-chapter-files/28276?nt=1 Status source · Retrieved Sep 14, 2026 - Modularity of elliptic curves over cyclotomic Z p-extensions of real quadratic fields https://arxiv.org/html/2206.12860v1 Status source · Retrieved Sep 14, 2026 - Modular elliptic curves and hyperbolic uniformization https://arxiv.org/abs/2607.03830 Status source · Retrieved Sep 14, 2026 Problem familyLanglands functoriality and modularity · 2 problems in this edition. Family membership alone does not assert an implication. - 7 · Global Langlands Functoriality Conjecture problem-global-langlands-functoriality-conjecture - 39 · Modularity of Elliptic Curves over Number Fields End K E =Z this problem Ranking uncertainty90% source-fit bootstrap rank range: 30–49. Source-fit opponent count: 43. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 39. Rank 040 Hilbert's Sixteenth Problem, Second Part For planar differential equations whose right-hand sides are polynomials of a fixed degree at least two, what is the maximum number of isolated closed trajectories, and how can they be arranged? Is that maximum finite for every degree?Probability, ergodic theory & dynamics Exact statementFor every degree n = 2, prove finiteness and determine the maximal number and possible configurations of limit cycles of planar polynomial vector fields of degree n. Why it mattersA resolution would determine how polynomial degree limits the number and arrangement of isolated periodic motions in planar systems. It would provide a complete degree-by-degree description of the possibilities across all vector fields in scope. Status Open Hilbert's sixteenth problem, part II, remains open. The checked purported quadratic bound faces a published critique; no universal bound for each degree was established. This is not a claim of a formal retraction absent such evidence. Status reviewed Sources - Luiz Fernando da Silva Gouveia and Joan Torregrosa, The local cyclicity problem. Melnikov method using Lyapunov constants, Proceedings of the Edinburgh Mathematical Society 65 2022 , 356-375, Introduction, p. 356. https://recercat.cat/bitstream/handle/2072/532013/MelnikovMethod.pdf?sequence=1 Statement source - A note on a recent attempt to solve the second part of Hilbert's 16th Problem https://arxiv.org/abs/2411.09594 Status source · Retrieved Sep 14, 2026 - Pesquisadores da Unesp propõem solução para desafio matemático em aberto há mais de um século https://jornal.unesp.br/2024/09/24/pesquisadores-da-unesp-propoem-solucao-para-desafio-matematico-em-aberto-ha-mais-de-um-seculo/ Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyHilbert's Sixteenth Problem, Second Part · 1 problem in this edition. Family membership alone does not assert an implication. - 40 · Hilbert's Sixteenth Problem, Second Part this problem Ranking uncertainty90% source-fit bootstrap rank range: 32–48. Source-fit opponent count: 44. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 40. Rank 041 The Quantum PCP Conjecture Even allowing a fixed small error per interaction, is estimating the lowest energy of a locally interacting quantum system as hard as every problem whose yes answers have a quantum proof that a quantum computer can check efficiently?Quantum information & computation Exact statementProve or disprove that approximating the ground-state energy of a local Hamiltonian to within a constant additive fraction of its number of local terms is QMA-hard. Why it mattersA proof would preserve QMA-hardness even at this coarse energy precision. A disproof would rule out the proposed constant-fraction hardness statement for the specified local-Hamiltonian problem. Status Open Quantum PCP remains open. The August 2026 classical private-PCP advance explicitly distinguishes the unresolved QMA problem and is not a quantum-PCP proof. Status reviewed Sources - Aharonov, Arad, and Vidick, The Quantum PCP Conjecture, Section 1.2, Conjecture 1.3 2013 . https://arxiv.org/html/1309.7495v1 Statement source - Private PCPs from Product Expansion https://eccc.weizmann.ac.il/report/2026/150/ Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyThe Quantum PCP Conjecture · 1 problem in this edition. Family membership alone does not assert an implication. - 41 · The Quantum PCP Conjecture this problem Ranking uncertainty90% source-fit bootstrap rank range: 34–49. Source-fit opponent count: 47. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 41. Rank 042 Conformal Field Theory Limit of the Critical Three-Dimensional Ising Model At criticality, does the three-dimensional Ising model have a conformally invariant continuum limit with non-Gaussian spin fluctuations? The question asks for both the symmetry of the limit and spin correlations that do not factorize as they would in a Gaussian field.Mathematical physics 3D Ising · 2 problems family-context-problem-conformal-field-theory-limit-of-the-critical-three-dimensional-ising-model Exact statementProve that the scaling limit of the critical three-dimensional Ising model exists and is conformally invariant, with a non-Gaussian spin field whose limiting spin correlations do not factorize according to Wick's rule. Why it mattersA proof would connect the critical lattice model to a conformally invariant continuum theory with non-Gaussian spin fluctuations. It would also establish the separately listed non-triviality target. Status Open The critical three-dimensional lattice Ising CFT/scaling-limit target remains open with verification limits. September fuzzy-sphere numerical certification concerns another model and does not prove convergence or conformal invariance of the frozen lattice limit. Status reviewed Sources - 100 Years of the Critical Ising Model on the Hypercubic Lattice Hugo Duminil-Copin . https://arxiv.org/html/2208.00864 Statement source - The Ising model: highlights and perspectives https://link.springer.com/article/10.1007/s11040-025-09515-1 Status source · Retrieved Sep 14, 2026 - New lower bounds for the near critical Ising and phi4 models' two-point functions https://www.unige.ch/~duminil/publi/2024 lower bound Ising.pdf Status source · Retrieved Sep 14, 2026 - Certified reduced-basis emulation of conformal field theories on the fuzzy sphere https://arxiv.org/abs/2609.11675 Status source · Retrieved Sep 14, 2026 Problem family3D Ising · 2 problems in this edition. Family membership alone does not assert an implication. - 42 · Conformal Field Theory Limit of the Critical Three-Dimensional Ising Model this problem - 74 · Non-Triviality of the Three-Dimensional Ising Model problem-non-triviality-of-the-three-dimensional-ising-model Ranking uncertainty90% source-fit bootstrap rank range: 30–49. Source-fit opponent count: 30. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 42. Rank 043 Grothendieck's Standard Conjecture D On a smooth projective variety, do rational algebraic cycles that give the same intersection numbers against every complementary cycle necessarily have the same class in every available Weil cohomology theory?Number theory & arithmetic geometry Grothendieck’s standard conjectures · 2 problems family-context-problem-grothendieck-standard-conjecture-d Exact statementFor every field k, every Weil cohomology theory H available over k, every smooth projective k-variety X of pure dimension n, every integer r with 0<=r<=n, and every rational algebraic cycle z of codimension r on X, if deg z.w =0 for every rational algebraic cycle w of codimension n-r on X, then the H cycle class of z is zero. Equivalently, numerical and H -homological equivalence of rational algebraic cycles agree in every codimension. Why it mattersA proof would make intersection numbers sufficient to detect homological vanishing in every codimension and cohomology theory in scope. A counterexample would separate numerical equivalence from homological equivalence. Status Open · subcases solved Standard Conjecture D remains open, with solved subcases. The August 2026 abelian-fourfold claim does not cover all smooth projective varieties; earlier conditional uses neither prove the universal target nor corroborate the newest claim. Status reviewed Sources - Michael K. Brown and Mark E. Walker, Standard Conjecture D for matrix factorizations, classical Conjecture1.1 and surrounding introduction. https://arxiv.org/pdf/1905.04626 Statement source - The Tate conjecture for abelian fourfolds over finite fields https://arxiv.org/abs/2608.28651 Status source · Retrieved Sep 14, 2026 - Tate's question, Standard conjecture D, semisimplicity and Dynamical degree comparison conjecture https://arxiv.org/abs/2503.09432 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyGrothendieck’s standard conjectures · 2 problems in this edition. Family membership alone does not assert an implication. - 43 · Grothendieck's Standard Conjecture D this problem - 93 · Grothendieck Standard Conjecture of Hodge Type problem-grothendieck-standard-conjecture-hodge-type Ranking uncertainty90% source-fit bootstrap rank range: 34–50. Source-fit opponent count: 39. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 43. Rank 044 The Exponential Time Hypothesis for 3-SAT The 3-SAT problem asks whether Boolean variables can satisfy constraints with three literals per clause. Is there a positive constant c such that no deterministic algorithm solves all n-variable instances in time 2^ cn ?Theoretical computer science Exact statementProve or disprove the Exponential Time Hypothesis: there exists a constant c 0 such that 3-SAT cannot be solved in deterministic time 2^{c n}. Why it mattersA proof would establish an exponential barrier to deterministic 3-SAT algorithms measured by variable count. A negative resolution would remove the asserted positive lower bound on the exponential rate. Status Open ETH remains open. The 2026 source treats positive-exponent hardness as a conjecture; conditional algorithmic lower bounds do not establish unrestricted exponential hardness. Status reviewed Sources - R. Impagliazzo, R. Paturi, J. Comput. System Sci. 62 2 :367-375, 2001. https://doi.org/10.1006/jcss.2000.1727 Statement source - Complexity Theory: Stronger forms of P = NP, Roei Tell https://www.cs.toronto.edu/~roei/2026%20complexity/6%20stronger%20lbs.pdf Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyThe Exponential Time Hypothesis for 3-SAT · 1 problem in this edition. Family membership alone does not assert an implication. - 44 · The Exponential Time Hypothesis for 3-SAT this problem Ranking uncertainty90% source-fit bootstrap rank range: 34–53. Source-fit opponent count: 45. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 44. Rank 045 Fourier restriction conjecture Do all the conjectured sharp norm bounds hold when a Fourier transform is restricted to the unit sphere, in every dimension at least two? These bounds control the transformed function’s size on the sphere using the original function’s size in space.Analysis & PDE Kakeya, restriction and smoothing estimates · 4 problems family-context-problem-fourier-restriction-conjecture Exact statementProve the conjectured sharp L^p R^d -to-L^q S^{d-1} boundedness range for restriction of the Fourier transform to the unit sphere in every dimension d at least 2, including the non-endpoint range specified by the standard Knapp necessary conditions. Why it mattersA proof would establish the stated sphere-restriction estimates across the proposed range of exponents. A counterexample at an admissible pair would reveal an additional obstruction beyond the stated necessary conditions. Status Open · subcases solved Fourier restriction remains open, with solved subcases. Sawyer's earlier full-proof version was withdrawn, while the current 2026 version treats averaged probabilistic estimates. Those do not settle the deterministic restriction conjecture. Status reviewed Sources - Diogo Oliveira e Silva, Communications in Mathematics 32 2024 https://cm.episciences.org/13778/pdf Statement source - Smooth Alpert frames, version 5 https://arxiv.org/abs/2311.03145v5 Status source · Retrieved Sep 14, 2026 - A probabilistic analogue of the Fourier extension conjecture https://arxiv.org/abs/2311.03145 Status source · Retrieved Sep 14, 2026 - A survey of Stein's restriction conjecture https://mathematics.stanford.edu/events/survey-steins-restriction-conjecture Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyKakeya, restriction and smoothing estimates · 4 problems in this edition. Family membership alone does not assert an implication. - 38 · Higher-dimensional Euclidean Kakeya conjecture problem-higher-dimensional-euclidean-kakeya-conjecture - 45 · Fourier restriction conjecture this problem - 202 · Bochner-Riesz conjecture problem-bochner-riesz-conjecture - 206 · Local Smoothing Conjecture for Wave Equations problem-local-smoothing-conjecture-for-wave-equations Ranking uncertainty90% source-fit bootstrap rank range: 38–53. Source-fit opponent count: 42. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 45. Rank 046 BQP versus QMA If a quantum computer can efficiently check a supplied quantum proof for a promised yes-or-no problem, can it also solve that problem efficiently without receiving the proof? Both verification and solving allow bounded error.Quantum information & computation Exact statementFor promise problems A= A yes,A no with disjoint sets of finite binary strings, use polynomial-time uniform quantum circuits over a fixed effectively specified finite universal gate set, with polynomially many gates and |0 ancillas and a measured output bit. BQP has such a circuit accepting each yes input with probability =2/3 and each no input with probability<=1/3. QMA permits a polynomial-length quantum witness: each yes input has a witness accepted with probability =2/3, and every witness on each no input is accepted with probability<=1/3. Outside the promise no condition is imposed. Is QMA contained in BQP, equivalently BQP=QMA? Why it mattersEquality would give efficient quantum decision procedures for QMA-complete promise problems, including the standard local-Hamiltonian energy-threshold problem. Separation would show that quantum witnesses add decision power beyond efficient quantum computation alone. Status Open BQP versus QMA remains open for unrelativized uniform promise classes. Unitary/classical oracle results and August qIO work with an assumed average-case QMA gap do not decide that equality. Status reviewed Sources - John Watrous, Quantum Computational Complexity; author-hosted survey,44pages; not attested identical to arXiv0804.3401. https://cs.uwaterloo.ca/~watrous/Papers/QuantumComputationalComplexity.pdf Statement source - CountCrypt: Quantum Cryptography between QCMA and PP https://eprint.iacr.org/2024/1707 Status source · Retrieved Sep 14, 2026 - Separating Quantum Indistinguishability Obfuscation from Falsifiable Assumptions https://arxiv.org/abs/2608.25195 Status source · Retrieved Sep 14, 2026 - Separating Quantum and Classical Advice with Good Codes https://arxiv.org/abs/2602.09385 Status source · Retrieved Sep 14, 2026 Problem familyBQP versus QMA · 1 problem in this edition. Family membership alone does not assert an implication. - 46 · BQP versus QMA this problem Ranking uncertainty90% source-fit bootstrap rank range: 38–54. Source-fit opponent count: 66. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 46. Rank 047 The Fontaine-Mazur Conjecture on Geometric Galois Representations A Galois representation records arithmetic symmetries as matrices. Must every irreducible l-adic representation over a number field, unramified outside finitely many places and de Rham at primes above l—a local condition describing behavior expected from geometry—come from a smooth projective variety’s cohomology, allowing subquotients and Tate twists?Number theory & arithmetic geometry Exact statementLet K be a number field and l a prime. Every irreducible continuous finite-dimensional l-adic representation rho of Gal Kbar/K , unramified outside finitely many places and de Rham at every place above l, occurs as a subquotient of H^i et X Kbar m for some smooth projective variety X over K, integer i = 0, and integer Tate twist m. Etale cohomology is taken with the same l-adic coefficient field as rho. There is no nonnegative Hodge-Tate weight restriction. Why it mattersA proof would make these arithmetic conditions sufficient for geometric realization across all number fields and finite dimensions. The realization may use any integer Tate twist, including negative twists, and imposes no nonnegative Hodge–Tate weight restriction. Status Open · subcases solved The full conjecture remains open, with solved subcases. The one-dimensional case is known; Thorne’s August 2026 preprint treats restricted two-dimensional cases and does not establish the assertion for every number field and dimension. This status is inferred from the reviewed scopes, not a certification of the proofs. Status reviewed Sources - Sug Woo Shin and Nicolas Templier, On fields of rationality for automorphic representations, author version dated 2 July 2014. https://math.berkeley.edu/~swshin/FieldRat.pdf Statement source - James Newton, Modularity of Galois Representations and Langlands Functoriality https://link.springer.com/article/10.1007/s41745-022-00305-0 Status source · Retrieved Sep 16, 2026 - Jack A. Thorne, Towards the Fontaine–Mazur conjecture for GL 2 https://arxiv.org/html/2608.07186v1 Status source · Retrieved Sep 16, 2026 Problem familyThe Fontaine-Mazur Conjecture on Geometric Galois Representations · 1 problem in this edition. Family membership alone does not assert an implication. - 47 · The Fontaine-Mazur Conjecture on Geometric Galois Representations this problem Ranking uncertainty90% source-fit bootstrap rank range: 38–55. Source-fit opponent count: 35. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 47. Rank 048 Nonexistence of Landau–Siegel zeros for quadratic Dirichlet L-functions Can real zeros of primitive quadratic Dirichlet L-functions be kept uniformly away from one on the logarithmic scale? For conductor q, the measure of the character’s arithmetic complexity, is there one positive constant δ excluding zeros between 1 − δ/log q and 1 for every q ≥ 3?Number theory & arithmetic geometry Riemann hypotheses and exceptional zeros · 4 problems family-context-problem-nonexistence-of-landau-siegel-zeros-for-quadratic-dirichlet-l-functions Exact statementDoes there exist an absolute constant δ 0 such that, for every primitive quadratic Dirichlet character χ of conductor q≥3, L β,χ ≠0 for every real β with 1−δ/log q<β<1? The logarithm is natural and the same δ must work for all conductors. Why it mattersA proof would remove the exceptional-real-zero possibility from this uniform region. It would eliminate that obstruction in applicable prime, class-number and L-value estimates while retaining their other hypotheses. Status Open The Landau–Siegel zero problem remains open. Recent special L-function and finite-range results are not universal Dirichlet zero exclusion. The September 14 review inspected Zhang v18; the related claim is now v19 dated September 14, so its new version was not independently adjudicated here and must not be represented as covered by the old assessment. Status reviewed Sources - Rick F. Lu, Asif Zaman and Haonan Zhao, Numerical Computations Concerning Landau–Siegel Zeros; arXiv2602.03626v1 February3,2026 https://arxiv.org/pdf/2602.03626v1 Statement source - On the Extended, Generalized, and Grand Riemann Hypotheses: A Unified Approach via General Properties of L-Functions, v18 https://www.preprints.org/manuscript/202506.0481/v18 Status source · Retrieved Sep 14, 2026 - Numerical Computations concerning Landau–Siegel Zeros https://arxiv.org/html/2602.03626v1 Status source · Retrieved Sep 14, 2026 - Landau-Siegel zeros of Rankin-Selberg L-functions https://arxiv.org/abs/2601.04189 Status source · Retrieved Sep 14, 2026 Problem familyRiemann hypotheses and exceptional zeros · 4 problems in this edition. Family membership alone does not assert an implication. - 2 · Riemann Hypothesis problem-riemann-hypothesis - 10 · Generalized Riemann Hypothesis for Dirichlet L-functions problem-the-generalized-riemann-hypothesis-for-dirichlet-l-functions - 11 · Extended Riemann hypothesis for Dedekind zeta functions problem-extended-riemann-hypothesis-for-dedekind-zeta-functions - 48 · Nonexistence of Landau–Siegel zeros for quadratic Dirichlet L-functions this problem Ranking uncertainty90% source-fit bootstrap rank range: 38–56. Source-fit opponent count: 52. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 48. Rank 049 Grothendieck Period Conjecture Periods encode integration data from algebraic geometry. For every Nori motive over the rationals, does the maximum number of algebraically independent periods equal the dimension of its motivic Galois group, which describes its algebraic symmetries?Number theory & arithmetic geometry Exact statementProve that for every Nori motive M over Q, trdeg Q Q periods of M equals dim G mot M , the dimension of its motivic Galois group. Why it mattersA proof would match the algebraic independence measured by the period field with the dimension of the motive’s symmetry group. A counterexample would show that those two quantities can differ. Status Open Grothendieck's arithmetic period conjecture remains open. Geometric period theorems and selected CM Kummer-surface results do not prove the all-motives algebraic-relations assertion. Status reviewed Sources - Formal statement, Conjecture 1.1 https://www.cambridge.org/core/journals/forum-of-mathematics-sigma/article/functional-transcendence-of-periods-and-the-geometric-andregrothendieck-period-conjecture/6B3E39BB9BC68DC2BC6E793D15B0735B Statement source - Functional Transcendence of Periods and the Geometric André--Grothendieck Period Conjecture https://arxiv.org/html/2208.05182v2 Status source · Retrieved Sep 14, 2026 - Grothendieck's period conjecture for Kummer surfaces of self-product CM type https://arxiv.org/abs/2303.05030 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyGrothendieck Period Conjecture · 1 problem in this edition. Family membership alone does not assert an implication. - 49 · Grothendieck Period Conjecture this problem Ranking uncertainty90% source-fit bootstrap rank range: 41–58. Source-fit opponent count: 46. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 49. Rank 050 Non-Perturbative Construction and Non-Triviality of 4D Euclidean Phi4 Quantum Field Theory Can a genuinely non-Gaussian scalar quantum field with quartic interaction be constructed in four Euclidean spacetime dimensions, satisfying the Osterwalder–Schrader axioms that make the Euclidean construction a rigorous quantum field theory?Mathematical physicsReviewed hold · conditional rank Exact statementConstruct a non-trivial, non-Gaussian Euclidean scalar quantum field theory with quartic interaction phi^4 4 in four spacetime dimensions satisfying the Osterwalder-Schrader axioms. Why it mattersA construction would realize the specified interacting field theory within those axioms. A general triviality theorem would instead exclude every such non-Gaussian construction in four dimensions. Status Reviewed hold · conditional rank Recorded hold reason. Retain open only with the inherited phi4 claim/scope warning. Accepted lattice-cutoff triviality is not a universal no-go theorem for every OS formulation. The separate nontrivial/OS-positivity construction claim lacks located independent acceptance; August mean-field-sector triviality is narrower. No positive or universal negative resolution is certified.The phi4 target remains open, with a claim and scope warning. Accepted lattice-cutoff triviality is not a universal no-go theorem for every OS formulation. Independent acceptance of the separate nontrivial/OS-positivity construction claim was not established; August mean-field-sector triviality is narrower. No positive or universal negative resolution is certified. Status reviewed Sources - M. Aizenman, H. Duminil-Copin, Ann. of Math. 194 1 :163-235, 2021. https://doi.org/10.4007/annals.2021.194.1.3 Statement source - Marginal triviality of the scaling limits of critical 4D Ising and phi4 models https://annals.math.princeton.edu/2021/194-1/p03 Status source · Retrieved Sep 14, 2026 - The non triviality of a Phi4 model, III, the Osterwalder-Schrader Positivity https://arxiv.org/abs/2101.11980 Status source · Retrieved Sep 14, 2026 - Triviality in a Non-Perturbative Second-Order Mean-Field Theory for phi4 https://arxiv.org/abs/2608.05998 Status source · Retrieved Sep 14, 2026 Problem familyNon-Perturbative Construction and Non-Triviality of 4D Euclidean Phi4 Quantum Field Theory · 1 problem in this edition. Family membership alone does not assert an implication. - 50 · Non-Perturbative Construction and Non-Triviality of 4D Euclidean Phi4 Quantum Field Theory this problem Ranking uncertainty90% source-fit bootstrap rank range: 44–61. Source-fit opponent count: 45. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 50. Rank 051 Equivalence of Worst-Case and Average-Case Hardness for NP Excluding Heuristica Suppose some efficiently verifiable decision problem resists polynomial-size circuits even when a separate circuit is chosen for each input length. Must some distributional NP problem—a problem together with its input distribution—also be hard on average, admitting no average-polynomial-time algorithm?Theoretical computer science Exact statementProve or disprove that if NP is hard in the worst case NP not in P/poly , then NP is hard on average DistNP not in AvgP . Why it mattersA proof would turn the stated worst-case circuit hardness assumption into average-case hardness. It would rule out the combination of that worst-case barrier with average-polynomial-time algorithms for every distributional NP problem. Status Open The exact NP not in P/poly implies DistNP not in AvgP target remains provisionally open. Located reductions add iO or quantitative hypotheses and do not recertify every stated nonuniform/AvgP convention; exact-formulation evidence remains limited. Status reviewed Sources - A. Bogdanov, L. Trevisan, SIAM J. Comput. 36 4 :1119-1159, 2006. https://doi.org/10.1137/050634629 Statement source - Cryptography meets worst-case complexity: Optimal security and more from iO and worst-case assumptions https://eccc.weizmann.ac.il/report/2025/072/download Status source · Retrieved Sep 14, 2026 - Hardness of Computing Nondeterministic Kolmogorov Complexity https://eccc.weizmann.ac.il/report/2025/203/ Status source · Retrieved Sep 14, 2026 Problem familyEquivalence of Worst-Case and Average-Case Hardness for NP Excluding Heuristica · 1 problem in this edition. Family membership alone does not assert an implication. - 51 · Equivalence of Worst-Case and Average-Case Hardness for NP Excluding Heuristica this problem Ranking uncertainty90% source-fit bootstrap rank range: 46–62. Source-fit opponent count: 38. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 51. Rank 052 The Baum-Connes Conjecture for the K-Theory of Reduced Group C -Algebras Can the topology of a group’s proper actions recover the K-theory invariants of its reduced operator algebra? The Baum–Connes conjecture asks whether the assembly map gives this exact match for every second-countable locally compact group.Algebra, representation & category theory Exact statementProve the Baum-Connes conjecture for every second-countable locally compact group G: the assembly map K top G → K C r G from the topological K-theory of the group, built from the classifying space for proper actions, to the K-theory of the reduced group C -algebra is an isomorphism. Why it mattersA proof would recover the specified operator-algebra K-theory from topological group data through this map. A counterexample would show either information lost by the map or target classes it cannot reach. Status Open Coefficient-free Baum–Connes remains open. Corrected coefficient formulations and failed weakened permanence hypotheses are different statements. Finn–Sell's lacunary-group preprint was withdrawn May 12, 2026 for a norm-control gap and is not accepted subcase support. Status reviewed Sources - W. Lück, H. Reich, in Handbook of K-theory, Springer, 2005. https://arxiv.org/abs/math/0402405 Statement source - There are no known counter-examples to the Baum-Connes conjecture https://www.fields.utoronto.ca/talks/There-are-no-known-counter-examples-to-Baum-Connes-conjecture Status source · Retrieved Sep 14, 2026 - Expanders and K-theory for group C algebras https://www.fields.utoronto.ca/talk-media/2/78/29/slides.pdf Status source · Retrieved Sep 14, 2026 - Controlled Analytic Properties and the Quantitative Baum-Connes Conjecture https://arxiv.org/abs/1908.02131 Status source · Retrieved Sep 14, 2026 - The Baum-Connes conjecture for extensions https://arxiv.org/abs/2508.05726 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyThe Baum-Connes Conjecture for the K-Theory of Reduced Group C -Algebras · 1 problem in this edition. Family membership alone does not assert an implication. - 52 · The Baum-Connes Conjecture for the K-Theory of Reduced Group C -Algebras this problem Ranking uncertainty90% source-fit bootstrap rank range: 48–62. Source-fit opponent count: 45. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 52. Rank 053 Schinzel's Hypothesis H on Simultaneous Prime Values of Irreducible Polynomials Can any fixed collection of irreducible integer polynomials with positive leading coefficients take prime values together at infinitely many positive inputs, provided no prime divides their product at every input? Irreducible means the polynomial cannot be factored further over the integers.Number theory & arithmetic geometry Simultaneous prime values · 5 problems family-context-problem-schinzel-s-hypothesis-h-on-simultaneous-prime-values-of-irreducible-polynomials Exact statementLet f1,...,fr in Z x be irreducible with positive leading coefficients and no prime dividing f1 n ...fr n for every n. Prove that infinitely many positive integers n make every fi n prime. Why it mattersA proof would make the absence of the stated divisibility obstruction sufficient for infinitely many simultaneous prime values. A counterexample would reveal a polynomial family with another obstruction to that conclusion. Status Open · subcases solved Schinzel H remains open, with solved subcases, with an unaccepted full-proof claim warning. Xiao's June 2025 posting matches the polynomial target but is explicitly unrefereed; later conditional use does not independently adjudicate it. Status reviewed Sources - Federico Scavia, Arithmetic Geometry and Algebraic Groups, Conjecture 3.1 2026 . https://www.lms.ac.uk/sites/default/files/inline-files/Scavia%20-%20Arithmetic%20Geometry%20and%20Algebraic%20Groups.pdf Statement source - On The Schinzel–Wójcik Problem Under Hypothesis H https://www.cambridge.org/core/journals/mathematical-proceedings-of-the-cambridge-philosophical-society/article/abs/on-the-schinzelwojcik-problem-under-hypothesis-h/C75A7C818639ADA780DE2A47C8BCFA19 Status source · Retrieved Sep 14, 2026 - From Golomb to Schinzel https://www.preprints.org/manuscript/202505.0651 Status source · Retrieved Sep 14, 2026 Problem familySimultaneous prime values · 5 problems in this edition. Family membership alone does not assert an implication. - 18 · Hardy-Littlewood Prime k-Tuple Conjecture problem-hardy-littlewood-prime-k-tuple-conjecture - 20 · Twin Prime Conjecture problem-twin-prime-conjecture - 53 · Schinzel's Hypothesis H on Simultaneous Prime Values of Irreducible Polynomials this problem - 58 · Dickson's Conjecture on Simultaneous Prime Values of Linear Forms problem-top500-omission-w045049-dickson-linear-prime-values - 81 · Bateman-Horn conjecture problem-bateman-horn-conjecture Ranking uncertainty90% source-fit bootstrap rank range: 46–66. Source-fit opponent count: 44. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 53. Rank 054 The Bombieri-Lang Conjecture for Varieties of General Type For every positive-dimensional variety of general type over a number field K, must its K-rational points lie in a proper algebraic subset, rather than being Zariski dense in the whole variety?Number theory & arithmetic geometry Exact statementProve that for every variety X of general type defined over a number field K, the set of K-rational points X K is not Zariski dense in X. Scope qualification. Interpret the inherited Bombieri-Lang target as the standard positive-dimensional general-type non-density conjecture over number fields. The zero-dimensional point is excluded; it is not a claimed counterexample to that canonical problem. The Campana positive implication is expressly limited to this standard scope. Why it mattersA proof would give a broad geometric reason that rational points cannot fill a variety. It would connect the general-type condition with arithmetic scarcity across all positive dimensions. Status Open Bombieri–Lang remains open for arithmetic non-density over number fields. The July 2026 specialist survey says so; the geometric function-field finite-to-abelian theorem does not resolve the arithmetic target. Status reviewed Sources - S. Lang, Bull. Amer. Math. Soc. 14 2 :159-205, 1986. https://doi.org/10.1090/S0273-0979-1986-15426-1 Statement source - Recent progress on the geometric Bombieri–Lang conjecture https://arxiv.org/html/2607.02165v1 Status source · Retrieved Sep 14, 2026 - A counting argument for the geometric Bombieri-Lang conjecture on ramified covers of abelian varieties https://arxiv.org/abs/2511.17010 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyThe Bombieri-Lang Conjecture for Varieties of General Type · 1 problem in this edition. Family membership alone does not assert an implication. - 54 · The Bombieri-Lang Conjecture for Varieties of General Type this problem Ranking uncertainty90% source-fit bootstrap rank range: 46–63. Source-fit opponent count: 54. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 54. Rank 055 Novikov Conjecture on homotopy invariance of higher signatures Higher signatures are numbers combining a manifold’s geometry with information about its fundamental group. For closed oriented smooth manifolds, do all rational higher signatures stay unchanged under orientation-preserving homotopy equivalence?Geometry & topology Exact statementFor every closed oriented smooth manifold M, every u in H^ B pi 1 M ; Q , and every classifying map r, prove that the higher signature <L M cup r^ u, M is invariant under orientation-preserving homotopy equivalence. Why it mattersA proof would make these higher signatures invariants of oriented homotopy type throughout the stated class of manifolds. A counterexample would give equivalent homotopy types with different higher-signature values. Status Open · subcases solved Novikov remains open, with solved subcases. Finite decomposition complexity and the August 2026 finite-complexity Banach-embedding criterion retain group restrictions. Embedding all groups in a Banach space does not remove the finite-complexity hypothesis. Status reviewed Sources - Eric Leichtnam and Paolo Piazza, Elliptic Operators and Higher Signatures, Question 1 2004 . https://aif.centre-mersenne.org/item/10.5802/aif.2049.pdf Statement source - Novikov's Conjecture https://arxiv.org/abs/1506.05408 Status source · Retrieved Sep 14, 2026 - Quantitative index, Novikov conjecture and coarse decomposability https://arxiv.org/abs/2412.01314 Status source · Retrieved Sep 14, 2026 - Embedding complexity into the universal Banach space and the strong Novikov conjecture https://arxiv.org/abs/2605.12930 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyNovikov Conjecture on homotopy invariance of higher signatures · 1 problem in this edition. Family membership alone does not assert an implication. - 55 · Novikov Conjecture on homotopy invariance of higher signatures this problem Ranking uncertainty90% source-fit bootstrap rank range: 49–62. Source-fit opponent count: 50. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 55. Rank 056 Equivariant Tamagawa number conjecture: Burns-Flach Conjecture4 package For a motive over a number field with extra symmetries, possibly noncommuting, does its equivariant L-function continue to s = 0, allowing poles, with the predicted zero or pole orders and rational and integral leading-term identities from its arithmetic cohomology? This is conditional on the Burns–Flach hypotheses on motivic cohomology, comparison maps, coefficient compatibility and coherence.Number theory & arithmetic geometry Exact statementFor every motive M over a number field K in Burns-Flach Section3.1's motivic-structure formalism, with an action of a finite-dimensional semisimple Q-algebra A, and every order mathcal A in A admitting the compatible projective mathcal A-structure of Section3.3, prove Burns-Flach Conjecture4 i - iv in its stated conditional setting. The external defining hypotheses are the finite motivic cohomology and real comparison exact sequence of Conjecture1, the motivic-to-p-adic comparison of Conjecture2 for M and Mdual 1 , coefficient compatibility giving well-defined equivariant local factors as in4.1 Conjecture3 or the weaker Euler-factor compatibility of Remark7 , and the stated Coherence hypothesis. A projective structure consists of full projective lattices in the archimedean realizations whose l-adic comparison images agree and are Galois-stable, not merely an arbitrary maximal-order choice. The INCLUDED assertions are: i the finite equivariant L-function L AM,s has meromorphic continuation to0; ii its componentwise integer order r equals rr A H f^1 K,Mdual 1 ^ -rr A H f^0 K,Mdual 1 ^ , with reduced rank as in2.6; iii for the nonzero componentwise leading Laurent coefficient L AM,0 =lim s- 0 s^ -r L AM,s , the class T Omega=delta hat^1 mathcal A,R L AM,0 +R Omega M,mathcal A lies in Cl mathcal A,Q ; and iv T Omega=0 in Cl mathcal A,R , hence in K 0 mathcal A,R . The R Omega term is the class of the fundamental virtual object with its integral compact-support cohomology structures and real comparison trivialization in3.4; delta hat is Lemma9's canonical extended reduced-norm boundary, not an inverse to a surjective real reduced norm. Equivalently the final identity is delta hat L =-R Omega. Preserve graded determinant conventions and all suitable nonmaximal/noncommutative orders. Coherence-free Remark8's Conjecture5 and Conjecture6 for EVERY prime are an alternative to the rational/integral clauses in the source's stated sense, not removal of i - ii or replacement by a single p-local or epsilon assertion. The external hypotheses are not presumed universally proved or restricted to currently established cases, and proving all of them is not added as a separate conclusion. Why it mattersUnder its stated hypotheses, the conjecture relates an equivariant L-function’s behavior at zero to arithmetic cohomology, including an integral leading-term identity. This relation must hold for every suitable order, including nonmaximal and noncommutative orders. Status Open · subcases solved The full Burns–Flach ETNC package remains open, with solved subcases. Recent conditional p-parts, modular motives and restricted Tate motives do not prove all four clauses for every motive and suitable order; specialist package-level currency remains qualified. Status reviewed Sources - David Burns and Matthias Flach, Tamagawa Numbers for Motives with Non-Commutative Coefficients, Documenta Mathematica6 2001 ,501-570 https://ems.press/content/serial-article-files/25892 Statement source - The Tamagawa number conjecture and Kolyvagin's conjecture for motives of modular forms https://link.springer.com/article/10.1007/s40687-026-00646-7 Status source · Retrieved Sep 14, 2026 - An unconditional proof of the abelian equivariant Iwasawa main conjecture and applications https://athene-forschung.unibw.de/92554?show id=155671 Status source · Retrieved Sep 14, 2026 Problem familyEquivariant Tamagawa number conjecture: Burns-Flach Conjecture4 package · 1 problem in this edition. Family membership alone does not assert an implication. - 56 · Equivariant Tamagawa number conjecture: Burns-Flach Conjecture4 package this problem Ranking uncertainty90% source-fit bootstrap rank range: 46–63. Source-fit opponent count: 66. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 56. Rank 057 Existence of One-Way Permutations Can one polynomial-time algorithm permute binary strings bijectively at every length n, while every randomized polynomial-time algorithm reverses a random output with success smaller than every inverse power of n for sufficiently large n?Cryptography, coding, information & optimization One-way functions and permutations · 2 problems family-context-problem-existence-of-one-way-permutations Exact statementThere exists a family f n:{0,1}^n- {0,1}^n, n =1, of bijections computed by one deterministic polynomial-time algorithm on 1^n,x , such that for every probabilistic polynomial-time algorithm A its inversion success Pr A 1^n,f n x =x , over uniform x and A's coins, is negligible in n: for every c 0 it is at most n^ -c for all sufficiently large n. This is unconditional existence in the classical unrelativized model, with no key sampling or variable domains. Why it mattersA proof would establish computationally hard inversion even when every output has exactly one preimage and the domain is all n-bit strings. The guarantee would hold uniformly across lengths in the classical model without oracles or sampled keys. Status Open Unconditional one-way permutations remain open. Full-domain trapdoor constructions require iO and one-way functions; query inversion taking square-root domain size is exponential in bit length and does not refute classical polynomial-time security. Status reviewed Sources - On Constructing One-Way Permutations from Indistinguishability Obfuscation, ePrint2015/752, TCC2016 version https://eprint.iacr.org/2015/752.pdf Statement source - On One-Shot Signatures, Quantum vs Classical Binding, and Obfuscating Permutations https://arxiv.org/abs/2507.12456 Status source · Retrieved Sep 14, 2026 - Quantum Search with In-Place Queries https://drops.dagstuhl.de/storage/00lipics/lipics-vol350-tqc2025/LIPIcs.TQC.2025.1/LIPIcs.TQC.2025.1.pdf Status source · Retrieved Sep 14, 2026 Problem familyOne-way functions and permutations · 2 problems in this edition. Family membership alone does not assert an implication. - 8 · Existence of One-Way Functions problem-existence-of-one-way-functions - 57 · Existence of One-Way Permutations this problem Ranking uncertainty90% source-fit bootstrap rank range: 44–67. Source-fit opponent count: 42. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 57. Rank 058 Dickson's Conjecture on Simultaneous Prime Values of Linear Forms For any finite family of integer linear forms with positive slopes, must infinitely many positive inputs make all values prime if no prime divides their product at every input?Number theory & arithmetic geometry Simultaneous prime values · 5 problems family-context-problem-top500-omission-w045049-dickson-linear-prime-values Exact statementFor every integer k =1 and every family f i n =a i n+b i 1<=i<=k with integers a i =1 and b i, if for each prime p some integer n has p not dividing the product of the f i n , then infinitely many positive integers n make every f i n prime. Why it mattersA proof would make the absence of these divisibility obstructions sufficient for infinitely many simultaneous prime values. A counterexample would reveal an additional obstruction for a particular family of linear forms. Status Open · subcases solved Dickson remains open, with solved subcases. May 2026 exposition explicitly retains the conjecture; Xiao's unaccepted Hypothesis H claim would imply the linear case, but no independent acceptance was established. Other inaccessible certificate/proof leads were not adopted. Status reviewed Sources - Shaohua Zhang, Notes on Dickson's Conjecture, arXiv:0906.3850v3 2009 https://arxiv.org/html/0906.3850 Statement source - An exposition on the supersimplicity of certain expansions of the additive group of the integers https://arxiv.org/html/2502.03915 Status source · Retrieved Sep 14, 2026 - From Golomb to Schinzel https://www.preprints.org/manuscript/202505.0651 Status source · Retrieved Sep 14, 2026 Problem familySimultaneous prime values · 5 problems in this edition. Family membership alone does not assert an implication. - 18 · Hardy-Littlewood Prime k-Tuple Conjecture problem-hardy-littlewood-prime-k-tuple-conjecture - 20 · Twin Prime Conjecture problem-twin-prime-conjecture - 53 · Schinzel's Hypothesis H on Simultaneous Prime Values of Irreducible Polynomials problem-schinzel-s-hypothesis-h-on-simultaneous-prime-values-of-irreducible-polynomials - 58 · Dickson's Conjecture on Simultaneous Prime Values of Linear Forms this problem - 81 · Bateman-Horn conjecture problem-bateman-horn-conjecture Ranking uncertainty90% source-fit bootstrap rank range: 45–69. Source-fit opponent count: 50. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 58. Rank 059 L = P Problem Can every decision problem solvable by a deterministic polynomial-time algorithm also be solved deterministically using only logarithmically many working-memory bits relative to its input length?Theoretical computer science Exact statementProve or disprove that every language decidable by a deterministic polynomial-time Turing machine is decidable using O log n work space; equivalently, determine whether the complexity classes L and P are equal. Why it mattersEquality would give logarithmic-workspace algorithms for every polynomial-time language. Separation would establish that some efficiently solvable decision problems require more working memory than that bound permits. Status Open L versus P remains open with source-quality and recency limits. Fixed polynomial time-space tradeoffs and square-root-space simulation do not give logarithmic space for every polynomial-time language. Status reviewed Sources - Cornell CS 6820 handout, The Circuit Value Problem, Fall 2022. https://www.cs.cornell.edu/courses/cs6820/2022fa/Handouts/CVP.pdf Statement source - Composing Low-Space Algorithms, revision 1 29 March 2026 https://eccc.weizmann.ac.il/report/2025/140/revision/1/download/ Status source · Retrieved Sep 14, 2026 - Simulating Time With Square-Root Space https://arxiv.org/abs/2502.17779 Status source · Retrieved Sep 14, 2026 Problem familyL = P Problem · 1 problem in this edition. Family membership alone does not assert an implication. - 59 · L = P Problem this problem Ranking uncertainty90% source-fit bootstrap rank range: 49–67. Source-fit opponent count: 40. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 59. Rank 060 Inverse Galois problem over the rational numbers Can every finite group occur as the Galois group—the group of field symmetries—of some finite Galois extension of the rational numbers?Algebra, representation & category theory Inverse Galois over Q · 2 problems family-context-problem-inverse-galois-problem-over-the-rational-numbers Exact statementProve or disprove that every finite group G is isomorphic to the Galois group Gal L/Q of some finite Galois extension L of the rational numbers Q. Why it mattersA proof would realize every finite abstract group as arithmetic symmetries over the rationals. A disproof would identify a finite group excluded from all such Galois extensions. Status Open · subcases solved The inverse Galois problem over Q remains open, with solved subcases. The August 2026 M23 preprint reports one finite group, not realization of every finite group; favorable informal commentary is not journal-proof certification. Status reviewed Sources - Fariba Ranjbar and Saeed Ranjbar, arXiv:1512.08708 2015 https://arxiv.org/abs/1512.08708 Statement source - The Mathieu group M 23 is a Galois group over Q https://arxiv.org/html/2608.08538v1 Status source · Retrieved Sep 14, 2026 - The inverse Galois challenge, part II https://galoisrepresentations.org/2026/08/16/the-inverse-galois-challenge-part-ii/ Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyInverse Galois over Q · 2 problems in this edition. Family membership alone does not assert an implication. - 60 · Inverse Galois problem over the rational numbers this problem - 116 · Regular Inverse Galois Problem over Q problem-regular-inverse-galois-problem-over-q Ranking uncertainty90% source-fit bootstrap rank range: 51–67. Source-fit opponent count: 48. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 60. Rank 061 NEXP versus P/poly If a decision problem’s yes answers have certificates whose length and verification time are at most exponential in a polynomial of input length, must polynomial-size Boolean circuits decide it, with a separate circuit chosen for every input length?Theoretical computer science Exact statementDetermine whether NEXP nondeterministic exponential time is contained in P/poly the class of languages computable by polynomial-size Boolean circuit families . Why it mattersContainment would give small circuits for every language in NEXP. Separation would establish that at least one nondeterministic exponential-time language requires circuit sizes beyond every polynomial bound. Status Open NEXP versus P/poly remains open on bounded evidence. Consistency of lower bounds with bounded arithmetic does not prove them in the standard model; the independent exact-status notes have no authenticated recent update date. Status reviewed Sources - Complexity Zoo: NEXP entry https://complexityzoo.net/Complexity Zoo:N Statement source - On the Consistency of Circuit Lower Bounds for Non-Deterministic Time https://arxiv.org/abs/2303.01016 Status source · Retrieved Sep 14, 2026 - From Godel incompleteness to the consistency of circuit lower bounds https://arxiv.org/abs/2604.25251 Status source · Retrieved Sep 14, 2026 - Basic Circuit Complexity — Chi-Ning Chou https://cnchou.github.io/notes/circuit.html Status source · Retrieved Sep 14, 2026 Problem familyNEXP versus P/poly · 1 problem in this edition. Family membership alone does not assert an implication. - 61 · NEXP versus P/poly this problem Ranking uncertainty90% source-fit bootstrap rank range: 52–73. Source-fit opponent count: 43. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 61. Rank 062 Permanent versus Determinant over C The permanent is the determinant-like sum with all permutation terms positive. If an n-by-n permanent is represented exactly as a determinant with affine-linear expressions in the original entries, must the required matrix size escape every polynomial bound in n, over the complex numbers?Theoretical computer science Algebraic computation lower bounds · 3 problems family-context-problem-permanent-versus-determinant-conjecture Exact statementOver C, let dc per n be the least m for which the n×n permanent per n X equals the determinant of an m×m matrix whose entries are affine-linear forms in the n^2 variables X, as an exact polynomial identity. Assert that dc per n is not O n^c for any real constant c 0: explicitly, for every c 0, C 0 and integer N≥1, some n≥N satisfies dc per n C n^c. No restrictions of equivariance, sparsity or symmetry are imposed on the representing matrix. This is exact, not border, determinantal complexity over C; the selected assertion is the source’s VP ws = VNP formulation, not an assertion that it is equivalent to VP = VNP. Why it mattersA proof would separate the permanent from polynomial-size weakly-skew algebraic computation. A polynomial-size determinant representation would refute this separation, even if no uniform algorithm for producing those representations were supplied. Status Open Permanent versus determinant remains open for unrestricted exact affine determinantal complexity over C. Quadratic lower bounds, border results and power-sum polynomials do not prove superpolynomial permanent growth; exact current-status evidence remains partly dated. Status reviewed Sources - Bürgisser, Ikenmeyer and Panova, arXiv:1604.06431v3, Conjecture 1.1; Landsberg and Ressayre, arXiv:1508.05788v2, Conjecture 1.1 and affine-linear representation definition. https://arxiv.org/pdf/1604.06431 Statement source - Bounds on Determinantal Complexity of Two Types of Generalized Permanents https://arxiv.org/html/2202.13016 Status source · Retrieved Sep 14, 2026 - A near-quadratic lower bound on the border determinantal complexity of sum i x i^n via conormal specialization https://arxiv.org/abs/2606.13628 Status source · Retrieved Sep 14, 2026 Problem familyAlgebraic computation lower bounds · 3 problems in this edition. Family membership alone does not assert an implication. - 32 · Valiants Algebraic P versus NP Conjecture VP = VNP problem-valiants-algebraic-p-versus-np-conjecture-vp-vnp - 62 · Permanent versus Determinant over C this problem - 340 · Koiran's Real Tau-Conjecture on Real Roots of Sums of Products of Sparse Polynomials problem-koiran-real-tau-conjecture Ranking uncertainty90% source-fit bootstrap rank range: 53–73. Source-fit opponent count: 57. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 62. Rank 063 Vojtas Main Conjecture on Height Inequalities in Diophantine Approximation Heights measure the arithmetic size of rational points. For every positive ε, can one remove a proper algebraic subset of a smooth projective variety over a number field so that proximity to a simple-normal-crossings divisor plus canonical height is bounded by ε times an ample height, up to an additive constant?Number theory & arithmetic geometry Exact statementLet k be a number field, X a smooth projective variety over k, K X a canonical divisor, A an ample divisor, D an effective simple normal-crossings divisor, and S a finite set of places containing the archimedean places. Prove that for every epsilon 0 there are a proper Zariski-closed subset Z of X and a constant C such that m {D,S} P +h {K X} P < epsilon h A P +C for every P in X k outside Z. Why it mattersA proof would uniformly control the stated combination of proximity and canonical height outside a proper algebraic exceptional set. A counterexample would require violations that cannot be confined by any permitted exceptional set and constant. Status Open Vojta's number-field height inequality remains open. Special integral-point results and function-field analogues change the quantifiers or ground field and do not establish every stated SNC-divisor inequality. Status reviewed Sources - Yasufuku, GCD inequalities inspired by Vojta's conjecture, Conjecture 8 2024 https://link.springer.com/article/10.1007/s00605-024-02018-1 Statement source - GCD inequalities inspired by Vojta's conjecture https://link.springer.com/article/10.1007/s00605-024-02018-1 Status source · Retrieved Sep 14, 2026 - Towards Lang--Vojta via Degeneration https://arxiv.org/abs/2602.06956 Status source · Retrieved Sep 14, 2026 - Lang-Vojta conjecture over function fields for very general log projective spaces https://arxiv.org/abs/2606.14074 Status source · Retrieved Sep 14, 2026 Problem familyVojtas Main Conjecture on Height Inequalities in Diophantine Approximation · 1 problem in this edition. Family membership alone does not assert an implication. - 63 · Vojtas Main Conjecture on Height Inequalities in Diophantine Approximation this problem Ranking uncertainty90% source-fit bootstrap rank range: 57–73. Source-fit opponent count: 44. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 63. Rank 064 Strong Exponential Time Hypothesis SETH As the allowed number of literals per Boolean clause grows, must the best satisfiability algorithms approach the brute-force exponent in 2^n, where n counts variables? The claim is that no fixed improvement of the exponent below one works for every fixed clause width.Theoretical computer science Exact statementProve or disprove that the optimal exponent s k for k-SAT satisfies lim {k to infinity} s k = 1. Why it mattersA proof would show that improvements in the exponential rate cannot stay uniformly below one as clause width increases. A disproof would establish a gap below that proposed limiting exponent. Status Open SETH remains open. Restricted resolution-over-parities lower bounds are not bounds against every deterministic k-SAT algorithm; 2026 scheduling work uses SETH as an assumption. Status reviewed Sources - C. Calabro, R. Impagliazzo, R. Paturi, J. Comput. System Sci. 74 3 :367-381, 2008. https://doi.org/10.1016/j.jcss.2007.07.007 Statement source - Strong ETH Holds for Bounded-Depth Resolution over Parities https://eccc.weizmann.ac.il/report/2025/188/ Status source · Retrieved Sep 14, 2026 - Tight S ETH-based Lower Bounds for Pseudopolynomial Algorithms for Bin Packing and Multi-Machine Scheduling https://arxiv.org/abs/2603.12999 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyStrong Exponential Time Hypothesis SETH · 1 problem in this edition. Family membership alone does not assert an implication. - 64 · Strong Exponential Time Hypothesis SETH this problem Ranking uncertainty90% source-fit bootstrap rank range: 60–74. Source-fit opponent count: 48. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 64. Rank 065 Artins Conjecture on the Holomorphy of Non-Abelian Artin L-Functions Does every Artin L-function attached to a nontrivial irreducible Galois representation extend to a holomorphic function on the entire complex plane, with no poles anywhere?Number theory & arithmetic geometry Exact statementProve that every non-trivial irreducible Artin L-function L s, rho associated to a Galois representation rho is an entire holomorphic function on the whole complex plane. Why it mattersA proof would establish entire continuation uniformly across the stated representation family. A counterexample would identify a nontrivial irreducible representation whose L-function fails that analytic property. Status Open Artin holomorphy remains open for all nontrivial irreducible Artin representations. The 2025 primary source and independent expert exposition retain the general question; special groups and criteria do not establish universal entireness. Status reviewed Sources - E. Artin, Zur Theorie der L-Reihen mit allgemeinen Gruppencharakteren, Abh. Math. Sem. Univ. Hamburg 8:292-306, 1930. https://doi.org/10.1073/pnas.16.8.532 Statement source - On holomorphy and non-vanishing of Artin L-functions https://link.springer.com/article/10.1007/s00605-025-02060-7 Status source · Retrieved Sep 14, 2026 - Artin's Conjectures — Ram Murty, Fields Institute, 12 March 2025 https://www1.fields.utoronto.ca/talks/Artins-Conjectures Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyArtins Conjecture on the Holomorphy of Non-Abelian Artin L-Functions · 1 problem in this edition. Family membership alone does not assert an implication. - 65 · Artins Conjecture on the Holomorphy of Non-Abelian Artin L-Functions this problem Ranking uncertainty90% source-fit bootstrap rank range: 59–76. Source-fit opponent count: 46. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 65. Rank 066 Modularity of Elliptic Curves over Totally Real Fields For an elliptic curve over a totally real number field, must there be a Hilbert cuspidal newform of parallel weight two with exactly the same L-function? Totally real means every embedding of the field into the complex numbers has real image.Number theory & arithmetic geometry Exact statementFor every totally real number field K and every elliptic curve E over K, prove that there is a Hilbert cuspidal newform f over K of parallel weight 2 satisfying L E,s =L f,s . Why it mattersA proof would match each elliptic curve in scope to an automorphic form through equality of their L-functions. A counterexample would identify a curve for which no form of the specified kind provides that match. Status Open · subcases solved Totally real modularity remains open, with solved subcases. The accessible theorem covers restricted towers with additional conditions, and finite database verification is not all curves over all fields. The final 2026 journal text could not be verified, so source currency remains qualified. Status reviewed Sources - Sho Yoshikawa, "On the modularity of elliptic curves over a composite field of some real quadratic fields," Research in Number Theory 2 2016 , article 31. https://arxiv.org/pdf/1607.05549 Statement source - Modularity of elliptic curves over cyclotomic Z p-extensions of real quadratic fields https://arxiv.org/html/2206.12860v1 Status source · Retrieved Sep 14, 2026 - Reliability of elliptic curve data over number fields reviewed https://www.lmfdb.org/knowledge/show/dq.ecnf.reliability Status source · Retrieved Sep 14, 2026 Problem familyModularity of Elliptic Curves over Totally Real Fields · 1 problem in this edition. Family membership alone does not assert an implication. - 66 · Modularity of Elliptic Curves over Totally Real Fields this problem Ranking uncertainty90% source-fit bootstrap rank range: 58–78. Source-fit opponent count: 54. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 66. Rank 067 The Farrell-Jones Conjecture in Algebraic K- and L-Theory Can the algebraic K- and L-theory of an integer group ring always be reconstructed from the data attached to its virtually cyclic subgroups—those containing a cyclic subgroup of finite index—through the Farrell–Jones assembly maps?Algebra, representation & category theory Exact statementProve the Farrell-Jones conjecture for every group G: the assembly maps from the equivariant homology of the classifying space for the family of virtually cyclic subgroups to the algebraic K-theory, respectively L-theory, of the group ring ℤG are isomorphisms. Why it mattersA proof would express both theories through the stated equivariant homology and assembly construction. A counterexample would locate a group where at least one map loses information or misses classes. Status Open · subcases solved Farrell–Jones remains open, with solved subcases. June 2026 primary work explicitly retains the general conjecture while proving specified group classes; broader coefficient frameworks do not remove group restrictions. Status reviewed Sources - Wolfgang Lück, Survey on the Farrell-Jones Conjecture, Conjectures 8.7 and 8.8, 2025. https://arxiv.org/html/2507.11337 Statement source - Automorphisms of relatively hyperbolic groups and the Farrell-Jones conjecture https://link.springer.com/article/10.1007/s00208-026-03431-7 Status source · Retrieved Sep 14, 2026 - On the Farrell-Jones conjecture for localising invariants https://arxiv.org/abs/2111.02490 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyThe Farrell-Jones Conjecture in Algebraic K- and L-Theory · 1 problem in this edition. Family membership alone does not assert an implication. - 67 · The Farrell-Jones Conjecture in Algebraic K- and L-Theory this problem Ranking uncertainty90% source-fit bootstrap rank range: 54–88. Source-fit opponent count: 24. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 67. Rank 068 Finiteness of the Tate-Shafarevich Group for Elliptic Curves For every elliptic curve over a number field, are there only finitely many classes in its Tate–Shafarevich group, which measures obstructions to turning local solutions into global ones?Number theory & arithmetic geometry Exact statementProve that the Tate-Shafarevich group Sha E/K is finite for every elliptic curve E defined over a number field K. Why it mattersA proof would establish finiteness separately for each elliptic curve and number field in scope. A counterexample would exhibit one with infinitely many Tate–Shafarevich classes; no uniform bound on all group sizes is requested. Status Open Sha finiteness for every elliptic curve over every number field remains open. The September 8 rank-two CM p-primary criterion and five-curve computations are not whole-group universal finiteness. A separate inaccessible proof-claim lead remains unverified. Status reviewed Sources - J. H. Silverman, The Arithmetic of Elliptic Curves, GTM 106, Springer, 2009. https://doi.org/10.1007/978-0-387-09494-6 Statement source - Rank stability in quadratic extensions and Hilbert’s tenth problem for the ring of integers of a number field https://link.springer.com/article/10.1007/s00222-025-01392-3 Status source · Retrieved Sep 14, 2026 - Finiteness of the Tate-Shafarevich group over function fields for groups of multiplicative type https://arxiv.org/abs/2607.15372 Status source · Retrieved Sep 14, 2026 - Second derivatives of p-adic L-functions and the Shafarevich–Tate group of rank-two CM elliptic curves https://arxiv.org/abs/2609.08431 Status source · Retrieved Sep 14, 2026 Problem familyFiniteness of the Tate-Shafarevich Group for Elliptic Curves · 1 problem in this edition. Family membership alone does not assert an implication. - 68 · Finiteness of the Tate-Shafarevich Group for Elliptic Curves this problem Ranking uncertainty90% source-fit bootstrap rank range: 56–85. Source-fit opponent count: 68. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 68. Rank 069 P versus NC^1 Can we exhibit a polynomial-time decision problem whose Boolean formulas built from AND, OR and NOT cannot have size bounded by any polynomial in the input length? Each length may have its own formula, but formulas cannot reuse intermediate results as circuits can.Theoretical computer science Exact statementExhibit a Boolean function family computable in deterministic polynomial time whose nonuniform De Morgan formula size is superpolynomial; equivalently, prove that P is not contained in nonuniform NC^1. Why it mattersA proof would show that polynomial-time computation cannot always be compressed into polynomial-size nonuniform formulas. It would separate P from containment in nonuniform NC¹ through an explicit kind of formula-size barrier. Status Open P versus nonuniform NC^1 remains open with dated-source qualification. Formula-composition progress does not establish the required superpolynomial De Morgan formula lower bound and is not the separate uniform P versus union-NC problem. Status reviewed Sources - Gavinsky, Meir, Weinstein, and Wigderson, ECCC TR13-190, revision 11, 2017. https://eccc.weizmann.ac.il/report/2013/190/revision/11/download Statement source - Toward Better Depth Lower Bounds: Strong Composition of XOR and a Random Function https://arxiv.org/abs/2410.10189 Status source · Retrieved Sep 14, 2026 - Toward Better Formula Lower Bounds: The Composition of a Function and a Universal Relation https://epubs.siam.org/doi/10.1137/15M1018319 Status source · Retrieved Sep 14, 2026 Problem familyP versus NC^1 · 1 problem in this edition. Family membership alone does not assert an implication. - 69 · P versus NC^1 this problem Ranking uncertainty90% source-fit bootstrap rank range: 57–87. Source-fit opponent count: 24. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 69. Rank 070 Derandomization of Polynomial Identity Testing Can a deterministic polynomial-time algorithm decide whether any given arithmetic circuit computes the zero polynomial, so that its output vanishes identically rather than just at selected inputs?Theoretical computer science Exact statementDetermine whether polynomial identity testing for arithmetic circuits deciding whether a given arithmetic circuit computes the identically zero polynomial can be derandomized into deterministic polynomial time. Why it mattersA positive answer would give an efficient identity test without random choices for every arithmetic circuit in scope. A negative answer would exclude every deterministic polynomial-time algorithm for this testing task. Status Open General arithmetic-circuit PIT derandomization remains open. July 2026 read-4 formula algorithms and nonassociative-algebra results do not cover arbitrary ordinary associative arithmetic circuits. Status reviewed Sources - Derandomizing Polynomial Identity over Finite Fields Implies Super-Polynomial Circuit Lower Bounds for NEXP Bin Fu https://arxiv.org/abs/1311.2358 Statement source - Polynomial Identity Testing for Read-4 Arithmetic Formulas https://drops.dagstuhl.de/storage/00lipics/lipics-vol383-ccc2026/LIPIcs.CCC.2026.25/LIPIcs.CCC.2026.25.pdf Status source · Retrieved Sep 14, 2026 - Efficient Polynomial Identity Testing Over Nonassociative Algebras https://arxiv.org/abs/2509.11349 Status source · Retrieved Sep 14, 2026 Problem familyDerandomization of Polynomial Identity Testing · 1 problem in this edition. Family membership alone does not assert an implication. - 70 · Derandomization of Polynomial Identity Testing this problem Ranking uncertainty90% source-fit bootstrap rank range: 60–83. Source-fit opponent count: 37. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 70. Rank 071 Deligne conjecture on coefficient-equivariant critical motivic L-values For a pure motive over the rationals, do its critical L-values, divided by the predicted periods and powers of 2πi, come from one element of its coefficient number field across every complex embedding? This asks whether geometric periods account for these values, assuming compatible realizations and analytic continuation giving finite values at those integers.Number theory & arithmetic geometry Exact statementFor every pure motive M over Q with coefficients in a number field E, in Deligne's geometric realization formalism Betti, de Rham and compatible l-adic realizations with comparison, dual, tensor and Tate-twist structures , and every critical integer n, the normalized finite L-value vector L M,n / 2 pi i ^ n d sign c^sign M lies in the diagonal E inside E tensor C, where sign= -1 ^n and d sign=dim E H B^sign M . Here H B^sign is the sign eigenspace of complex conjugation, and c^sign is the determinant of the Betti/de Rham comparison to the corresponding filtration quotient, computed in E-rational bases and defined modulo E-units as in Deligne1.7/2.6. Critical means that L infinity sigma,M,s and L infinity sigma,Mdual,1-s have no pole at s=n for every embedding sigma:E- C equivalently use Mdual 1 at -s . Use finite, not completed, L-functions with the compatible E-valued local factors of2.2. Analytic continuation giving finite values at the selected critical integer and the source's well-defined compatible realizations/local factors are defining prerequisites, not assertions proved by this period law and not a restriction to cases whose continuation is already known. The simultaneous embedding vector must come from one element of E; unrelated componentwise algebraicity is insufficient. All-zero values are allowed, but mixed zero/nonzero components do not satisfy the diagonal membership. No general nonvanishing or order-of-zero assertion from Deligne2.7 is appended. Why it mattersA proof would relate critical L-values to Betti–de Rham periods compatibly across all coefficient embeddings. The requirement is one shared algebraic element; separate algebraicity of unrelated components would not suffice. Status Open · subcases solved Deligne critical values remain open, with solved subcases for the coefficient-equivariant universal motive target. Recent CM-automorphic results retain regularity/rationality hypotheses; numerical checks on curves are not universal proofs. Status reviewed Sources - Kazuki Morimoto, On special values of L-functions for quaternion unitary groups of degree2 and GL 2 , Oberwolfach Report22/2014; with Deligne's definitions and twist convention https://ems.press/content/serial-article-files/46514?nt=1 Statement source - Factorization of periods, construction of automorphic motives and Deligne's conjecture over CM-fields https://arxiv.org/abs/2509.02303 Status source · Retrieved Sep 14, 2026 - Motivic pieces of curves: L-functions and periods https://arxiv.org/abs/2601.21934 Status source · Retrieved Sep 14, 2026 Problem familyDeligne conjecture on coefficient-equivariant critical motivic L-values · 1 problem in this edition. Family membership alone does not assert an implication. - 71 · Deligne conjecture on coefficient-equivariant critical motivic L-values this problem Ranking uncertainty90% source-fit bootstrap rank range: 61–77. Source-fit opponent count: 53. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 71. Rank 072 Termination of Flips Conjecture in the Minimal Model Program The minimal model program simplifies algebraic varieties through controlled transformations called flips. For any projective log canonical pair over the complex numbers, must every sequence of flips end after finitely many steps, in every dimension?Geometry & topology Minimal model conjectures · 3 problems family-context-problem-termination-of-flips-conjecture-in-the-minimal-model-program Exact statementProve that there is no infinite sequence of flips in the minimal model program: for a projective log canonical or klt pair over the complex numbers, every sequence of K X + Delta -flips terminates after finitely many steps, in every dimension. Why it mattersA proof would guarantee that the flip stage cannot continue indefinitely, regardless of which permitted sequence is chosen. This would remove infinite flipping as an obstruction in the stated minimal-model setting in every dimension. Status Open · subcases solved Termination of flips remains open, with solved subcases. The 2026 Enriques theorem is restricted, and other all-sequence or dimension-four implications retain unproved hypotheses; none proves arbitrary flip-sequence termination in all dimensions. Status reviewed Sources - Vladimir Lazic, Zhixin Xie, "Nakayama-Zariski decomposition and the termination of flips," arXiv:2305.01752v4; received July 8, 2024, accepted June 25, 2025. https://master-math-fonda.imj-prg.fr/gt/mmp.pdf Statement source - MMP for Enriques pairs and singular Enriques varieties https://jep.centre-mersenne.org/item/10.5802/jep.334.pdf Status source · Retrieved Sep 14, 2026 - Nakayama-Zariski decomposition and the termination of flips https://epiga.episciences.org/16998 Status source · Retrieved Sep 14, 2026 - On termination of flips and exceptionally non-canonical singularities https://arxiv.org/abs/2209.13122 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyMinimal model conjectures · 3 problems in this edition. Family membership alone does not assert an implication. - 72 · Termination of Flips Conjecture in the Minimal Model Program this problem - 78 · Abundance Conjecture for the minimal model program problem-abundance-conjecture-for-the-minimal-model-program - 193 · Nonvanishing conjecture for projective log canonical pairs problem-nonvanishing-conjecture-for-projective-log-canonical-pairs Ranking uncertainty90% source-fit bootstrap rank range: 63–88. Source-fit opponent count: 30. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 72. Rank 073 FPT versus W 1 Can we deterministically decide whether an n-vertex graph contains k mutually connected vertices in time f k n^c, for some computable function f and fixed constant c independent of k? This would confine any non-polynomial growth in the bound to the chosen clique size k.Theoretical computer science Exact statementProve or disprove FPT =W 1 in classical uniform parameterized complexity. Equivalently, determine whether there is a deterministic algorithm, a total computable function f:N- N and an absolute constant c such that on every finite simple n-vertex graph G and every integer 1<=k<=n it decides the existence of a k-clique within f k n^c steps. The conjecture asserts no such algorithm. The function and exponent are fixed with the algorithm; no input-length advice, oracle or nonuniform circuit family is allowed. Scope qualification. FPT = W 1 would imply P = NP: P = NP would give a polynomial-time Clique decider, which is an allowed FPT decider. The equality outcome FPT = W 1 allows arbitrary computable parameter dependence and does not itself supply a polynomial-time Clique algorithm in the full input size. The completion criterion remains the stated uniform parameterized class question; no additional theorem is a separate requirement. Why it mattersProving that no such uniform algorithm exists would separate FPT from W 1 and imply P ≠ NP. An FPT algorithm would allow arbitrary computable dependence on k, so it would not by itself give polynomial running time in the full input size. Status Open FPT versus W 1 remains open. Yi's counting-collapse claim was explicitly withdrawn May 10, 2026; independent 2026 work retains the separation as an assumption. The ordinary uniform k-Clique target is unchanged. Status reviewed Sources - Chen, Feng, Laekhanukit and Liu, Simple Combinatorial Construction, arXiv:2304.07516v2 2024 . https://arxiv.org/html/2304.07516v2 Statement source - Sharp-W\ 1\ = FPT: Fixed-Parameter Tractable Exact Algorithms for the k-Matching Problem https://arxiv.org/abs/2604.16308 Status source · Retrieved Sep 14, 2026 - A Parameterized-Complexity Framework for Finding Local Optima https://drops.dagstuhl.de/storage/00lipics/lipics-vol362-itcs2026/LIPIcs.ITCS.2026.66/LIPIcs.ITCS.2026.66.pdf Status source · Retrieved Sep 14, 2026 Problem familyFPT versus W 1 · 1 problem in this edition. Family membership alone does not assert an implication. - 73 · FPT versus W 1 this problem Ranking uncertainty90% source-fit bootstrap rank range: 62–90. Source-fit opponent count: 43. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 73. Rank 074 Non-Triviality of the Three-Dimensional Ising Model Does the large-scale limit of the critical three-dimensional Ising spin model remain non-Gaussian, with correlations that do not reduce to sums of products of pair correlations as Wick’s rule would require?Mathematical physics 3D Ising · 2 problems family-context-problem-non-triviality-of-the-three-dimensional-ising-model Exact statementProve that the critical three-dimensional Ising model is non-trivial, that is, that its scaling limit is not Gaussian and its critical spin correlations do not factorize according to Wick's rule. Why it mattersA proof would establish that the limiting correlations retain behavior beyond a Gaussian theory. A Gaussian limit with Wick factorization would refute the stated non-triviality claim. Status Open Three-dimensional short-range Ising nontriviality remains provisionally open with an exact-proof warning. Long-range effective-dimension-four Gaussianity is a different model. The accessible Kai exponent-certificate abstract and demo/surrogate artifact do not establish the frozen non-Wick lattice scaling-limit theorem; full article/artifact correspondence was not verified. Status reviewed Sources - Lectures on the Ising and Potts models on the hypercubic lattice Hugo Duminil-Copin . https://arxiv.org/html/1707.00520 Statement source - Triviality of the scaling limits of critical Ising and phi^4 models with effective dimension at least four https://arxiv.org/abs/2309.05797 Status source · Retrieved Sep 14, 2026 - Lectures on the Ising and Potts models on the hypercubic lattice https://www.ihes.fr/~/duminil/publi/2017PIMS.pdf Status source · Retrieved Sep 14, 2026 - Certified three-dimensional Ising critical exponents from harmonized bootstrap series and interval Monte Carlo https://link.springer.com/article/10.1140/epjb/s10051-025-01109-8 Status source · Retrieved Sep 14, 2026 - Certified Three Dimensional Ising Critical Exponents: author release notes https://www.researchgate.net/publication/397730833 Certified Three Dimensional Ising Critical Exponents Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem family3D Ising · 2 problems in this edition. Family membership alone does not assert an implication. - 42 · Conformal Field Theory Limit of the Critical Three-Dimensional Ising Model problem-conformal-field-theory-limit-of-the-critical-three-dimensional-ising-model - 74 · Non-Triviality of the Three-Dimensional Ising Model this problem Ranking uncertainty90% source-fit bootstrap rank range: 66–87. Source-fit opponent count: 32. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 74. Rank 075 Montgomerys Pair Correlation Conjecture for Zeros of the Riemann Zeta Function If vertical separations between pairs of nontrivial zeta zeros are rescaled by the local mean spacing of zeros, does their limiting pair density at every fixed separation u 0 equal 1 − sin πu / πu ²?Number theory & arithmetic geometry Exact statementProve that normalized vertical gaps between pairs of nontrivial zeros of the Riemann zeta function have limiting pair density 1 - sin pi u / pi u ^2 at fixed positive normalized separation. Why it mattersA proof would give the specified quantitative law for two-point spacing statistics of zeta zeros. A counterexample would show a departure from that density at a positive normalized separation. Status Open Montgomery pair correlation remains open. Jerby's December 2025 proof claim assumes RH, which the stated target does not assume, and lacks established independent acceptance. It is recorded as an unverified conditional claim, not an accepted conditional theorem or an unconditional resolution. Status reviewed Sources - Goldston, Lee, Schettler, and Suriajaya, Pair Correlation Conjecture for the zeros of the Riemann zeta-function I version 4, 2026 , Section 4, PCC. https://arxiv.org/html/2503.15449v4 Statement source - Variations of the Hardy Z-Function and the Montgomery Pair Correlation Conjecture https://arxiv.org/abs/2511.18275 Status source · Retrieved Sep 14, 2026 - Pair correlation of zeros of Dirichlet L-functions: a possible path towards the conjectures of Chowla, Elliott-Halberstam and Montgomery https://link.springer.com/article/10.1007/s00208-026-03383-y Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyMontgomerys Pair Correlation Conjecture for Zeros of the Riemann Zeta Function · 1 problem in this edition. Family membership alone does not assert an implication. - 75 · Montgomerys Pair Correlation Conjecture for Zeros of the Riemann Zeta Function this problem Ranking uncertainty90% source-fit bootstrap rank range: 63–96. Source-fit opponent count: 34. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 75. Rank 076 Woodins Ultimate L Conjecture for Supercompact Cardinals If an extendible cardinal δ exists, must there be an inner model of set theory inside HOD—the hereditarily ordinal-definable sets—that satisfies the Ultimate-L axiom and is a weak extender model for δ’s supercompactness?Logic, foundations & set theory Exact statementAssuming delta is an extendible cardinal, prove that there is an inner model N contained in HOD such that N is a weak extender model for the supercompactness of delta and N satisfies V=Ultimate-L. Why it mattersAn inner model is an internal universe of sets; HOD consists of hereditarily ordinal-definable sets. The conjecture would build such a universe inside HOD that retains the specified supercompactness structure and satisfies Ultimate-L under the ambient extendibility assumption. Status Open Ultimate-L remains provisionally open with a formulation/consistency warning. The exacting-cardinal obstruction is conditional, not an unconditional disproof. Woodin's August 26, 2026 formulation remains conjectural but differs from the stated weak-extender-inside-HOD wording; equivalence requires specialist adjudication. The unrefereed broad proof essay was not accepted. Status reviewed Sources - Bagaria and Ternullo, Steel's Programme: Evidential Framework, the Core and Ultimate-L, Review of Symbolic Logic 2021 , Conjecture 5.2. https://www.cambridge.org/core/journals/review-of-symbolic-logic/article/steels-programme-evidential-framework-the-core-and-ultimatel/22BC5068C2DC48CC899C60E1445A7792 Statement source - Large cardinals, structural reflection, and the HOD Conjecture https://arxiv.org/abs/2411.11568 Status source · Retrieved Sep 14, 2026 - The AD+ duality program, the HOD conjecture, and the Ultimate-L conjecture https://link.springer.com/content/pdf/10.1007/s11537-026-2541-4.pdf Status source · Retrieved Sep 14, 2026 - Proof of the Omega-Conjecture and Establishment of the V = Ultimate L Axiom: Mathematical Foundations and Physical Realization https://www.preprints.org/frontend/manuscript/f77017e8752bef0a5fb615eb22525dd8/download pub Status source · Retrieved Sep 14, 2026 Problem familyWoodins Ultimate L Conjecture for Supercompact Cardinals · 1 problem in this edition. Family membership alone does not assert an implication. - 76 · Woodins Ultimate L Conjecture for Supercompact Cardinals this problem Ranking uncertainty90% source-fit bootstrap rank range: 65–97. Source-fit opponent count: 33. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 76. Rank 077 Chowla conjecture on correlations of the Liouville function The Liouville function is +1 or −1 according to whether an integer has an even or odd number of prime factors, counted with multiplicity. For k ≥ 2 linear forms a i n + b i with natural coefficients a i, distinct nonnegative offsets b i and no proportional pair, do products of these signs average to zero?Probability, ergodic theory & dynamics Exact statementProve that for the Liouville function lambda, for every k = 2, distinct nonnegative integers b 1,...,b k and natural numbers a 1,...,a k with a i b j - a j b i = 0 for i<j, sum {n<=x} lambda a 1 n + b 1 ... lambda a k n + b k = o x as x - infinity. Why it mattersThe Liouville function is +1 or −1 according to whether the number of prime factors, counted with multiplicity, is even or odd. A proof would show that positive and negative products balance in the limit for every fixed admissible family of affine forms. Status Open Chowla remains open for every stated nonproportional affine-form tuple. Xiao's unrefereed fixed-shift claim is overlapping but not explicitly all affine forms; logarithmic and averaged August results do not establish every ordinary Cesaro correlation. Status reviewed Sources - Terence Tao, Equivalence of the logarithmically averaged Chowla and Sarnak conjectures, abstract 2016 . https://arxiv.org/abs/1605.04628 Statement source - The Chowla conjecture and Landau–Siegel zeroes https://www.cambridge.org/core/journals/mathematical-proceedings-of-the-cambridge-philosophical-society/article/chowla-conjecture-and-landausiegel-zeroes/515F3378450DED201A21E26BF801CEEB Status source · Retrieved Sep 14, 2026 - From Bateman-Horn to Chowla https://www.preprints.org/manuscript/202601.0277 Status source · Retrieved Sep 14, 2026 - Two averaged dynamical generalizations of Chowla's conjecture https://arxiv.org/abs/2608.16108 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyChowla conjecture on correlations of the Liouville function · 1 problem in this edition. Family membership alone does not assert an implication. - 77 · Chowla conjecture on correlations of the Liouville function this problem Ranking uncertainty90% source-fit bootstrap rank range: 70–95. Source-fit opponent count: 35. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 77. Rank 078 Abundance Conjecture for the minimal model program A nef divisor has nonnegative degree on every curve. For a minimal projective pair with Kawamata log terminal singularities, must some positive multiple of its canonical-plus-boundary divisor have enough global sections to define a map everywhere, with no common zero?Geometry & topology Minimal model conjectures · 3 problems family-context-problem-abundance-conjecture-for-the-minimal-model-program Exact statementProve that for every minimal projective Kawamata log terminal pair X,Delta , the nef adjoint divisor K X+Delta is semiample. Why it mattersA proof would turn the nefness condition into semiampleness, producing the associated pluricanonical morphism. It would let every minimal pair in scope determine a geometric map through a suitable multiple of its adjoint divisor. Status Open Abundance remains open, with known restricted cases. Chern-class equality cases, boundary-restricted results and reductions of Kaehler abundance to projective abundance do not prove semiampleness for every stated projective klt pair. Status reviewed Sources - Vladimir Lazic, Metrics with minimal singularities and the Abundance conjecture 2024 https://arxiv.org/abs/2406.18233 Statement source - Inequalities of Miyaoka-Yau type and Uniformisation of varieties of intermediate Kodaira Dimension https://arxiv.org/abs/2601.15138 Status source · Retrieved Sep 14, 2026 - An approach to the abundance conjecture for Kaehler varieties via algebraic reduction https://arxiv.org/abs/2604.03879 Status source · Retrieved Sep 14, 2026 - Two non-vanishing results concerning the anti-canonical bundle https://www.cambridge.org/core/journals/nagoya-mathematical-journal/article/two-nonvanishing-results-concerning-the-anticanonical-bundle/1E68F2B88B818A7CBD47B96B7D700548 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyMinimal model conjectures · 3 problems in this edition. Family membership alone does not assert an implication. - 72 · Termination of Flips Conjecture in the Minimal Model Program problem-termination-of-flips-conjecture-in-the-minimal-model-program - 78 · Abundance Conjecture for the minimal model program this problem - 193 · Nonvanishing conjecture for projective log canonical pairs problem-nonvanishing-conjecture-for-projective-log-canonical-pairs Ranking uncertainty90% source-fit bootstrap rank range: 68–97. Source-fit opponent count: 41. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 78. Rank 079 Graph Isomorphism in Polynomial Time Can a deterministic classical algorithm decide in polynomial time whether any two finite graphs describe the same connections after their vertices are relabeled?Theoretical computer science Exact statementDetermine whether the graph isomorphism problem for general finite graphs is decidable by a deterministic classical Turing machine in polynomial time. Why it mattersA positive answer would give an efficient general procedure for recognizing identical graph structure despite different labels. A negative answer would establish a polynomial-time barrier even for this precisely defined comparison task. Status Open General graph isomorphism in deterministic polynomial time remains open. Heated-space methods explicitly lack all-input guarantees; the older Xue claim was withdrawn after an author-reported counterexample. A separate inaccessible 2026 claim was not accepted. Status reviewed Sources - E. W. Weisstein, 'Graph Isomorphism,' MathWorld, accessed 2026. https://mathworld.wolfram.com/GraphIsomorphism.html Statement source - Graph Isomorphism: Mixed-Integer Convex Optimization from First-Order Methods https://arxiv.org/abs/2512.17417 Status source · Retrieved Sep 14, 2026 - Finding Graph Isomorphisms in Heated Spaces in Almost No Time https://arxiv.org/abs/2601.03787 Status source · Retrieved Sep 14, 2026 - Testing Isomorphism of Graphs in Polynomial Time https://arxiv.org/abs/2305.12688 Status source · Retrieved Sep 14, 2026 Problem familyGraph Isomorphism in Polynomial Time · 1 problem in this edition. Family membership alone does not assert an implication. - 79 · Graph Isomorphism in Polynomial Time this problem Ranking uncertainty90% source-fit bootstrap rank range: 67–104. Source-fit opponent count: 29. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 79. Rank 080 Hadwiger Conjecture If a graph has no complete graph on t + 1 vertices as a minor, can its vertices always be colored with t colors so adjacent vertices have different colors, for every positive integer t? A minor is obtained by deleting vertices or edges and contracting edges.Combinatorics & discrete geometry Exact statementProve that every simple graph with no K {t+1} minor is t-colourable for every integer t = 1. Why it mattersA proof would make exclusion of a complete minor sufficient to control chromatic number. It would connect a graph’s coloring requirements with the complete structures obtainable by deleting and contracting its edges. Status Open Ordinary Hadwiger remains open. The September 9 Liu–Luo preprint improves an approximate chromatic bound, not the exact t-1 threshold; odd-minor counterexamples concern a stronger variant. October 6, 2026: Norin and Steiner 4 October claim Ct-colourability for K t-minor-free graphs, with an absolute C, and credit GPT-6 Astra. This linear bound does not establish the sharp t-1 colours required by ordinary Hadwiger. The complete proof was not audited in this review. Status reviewed Sources - Lafferty and Song, Every graph with no K 8^{-4} minor is 7-colorable 2022 , abstract. https://arxiv.org/abs/2208.07338 Statement source - Beyond halfway to Hadwiger's conjecture https://arxiv.org/abs/2609.06867v2 Status source · Retrieved Sep 14, 2026 - Disproof of the Odd Hadwiger Conjecture https://arxiv.org/abs/2512.20392 Status source · Retrieved Sep 14, 2026 - Hadwiger Conjecture: October 2026 scope review https://arxiv.org/abs/2610.05291 Status source · Retrieved Oct 6, 2026 Research on ProofAtlas Problem familyHadwiger Conjecture · 1 problem in this edition. Family membership alone does not assert an implication. - 80 · Hadwiger Conjecture this problem Ranking uncertainty90% source-fit bootstrap rank range: 69–97. Source-fit opponent count: 35. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 80. Rank 081 Bateman-Horn conjecture For a fixed collection of distinct irreducible integer polynomials with positive leading coefficients and no prime dividing their product at every integer, does the Bateman–Horn formula correctly predict how often they are all prime at the same positive input, including the leading constant from their divisibility patterns?Number theory & arithmetic geometry Simultaneous prime values · 5 problems family-context-problem-bateman-horn-conjecture Exact statementProve or disprove the Bateman-Horn asymptotic formula for the number of positive integers n up to x for which any fixed admissible finite family of distinct irreducible integer polynomials with positive leading coefficients are simultaneously prime. Why it mattersA proof would quantify how frequently simultaneous polynomial prime values occur, including the constant prescribed for each family. It would supply counting predictions more precise than merely asserting infinitely many successful inputs. Status Open · subcases solved Bateman–Horn remains open, with solved subcases with an unaccepted full-proof claim warning. Xiao's posting invokes hypothesis F and is unrefereed; a 2026 random-set model is not a theorem for the actual primes. Status reviewed Sources - S. L. Aletheia-Zomlefer, L. Fukshansky, and S. R. Garcia, Expositiones Mathematicae 38 2020 , 430-479 https://www.sciencedirect.com/science/article/am/pii/S0723086918301178 Statement source - From Golomb to Bateman-Horn https://www.preprints.org/manuscript/202512.2666 Status source · Retrieved Sep 14, 2026 - Sets of integers satisfying Bateman-Horn statistics https://arxiv.org/abs/2605.01155 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familySimultaneous prime values · 5 problems in this edition. Family membership alone does not assert an implication. Ranking uncertainty90% source-fit bootstrap rank range: 67–103. Source-fit opponent count: 27. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 81. Rank 082 The Lindeloef Hypothesis on the Growth of the Riemann Zeta Function As we move up or down the line with real part one-half, is the magnitude of the Riemann zeta function eventually bounded by every positive power of the height above or below the real axis, up to a constant depending on that power?Number theory & arithmetic geometry Exact statementProve that for every epsilon 0 there is a constant C epsilon such that |zeta 1/2+it | <= C epsilon |t|^epsilon for all sufficiently large real |t|. Why it mattersA proof would rule out growth at any fixed positive power along the critical line, controlling even its unusually large values. A disproof would demonstrate repeated growth exceeding at least one such power bound. Status Open Lindelof remains open with a renewed full-proof claim warning. Unterberger's v48 was withdrawn August 12, then v49 and September 4 v50 restored the claim. No independent assessment of the repair was established; it must not be called simply withdrawn, accepted or refuted. A separate formula argument explicitly lacks convergence justification. Status reviewed Sources - Alexandra Florea, A survey of moment bounds for zeta s : From Heath-Brown's work to the present, Journal of the London Mathematical Society 2026 . https://londmathsoc.onlinelibrary.wiley.com/doi/10.1112/jlms.70376 Statement source - Analogues of the Lindelöf Hypothesis for the Barnes multiple zeta function and related problems https://arxiv.org/abs/2606.02407 Status source · Retrieved Sep 14, 2026 - An approach to the Lindelöf Hypothesis for Dirichlet L-functions https://arxiv.org/abs/2602.05731 Status source · Retrieved Sep 14, 2026 - Pseudodifferential arithmetic, Riemann and Lindelöf hypotheses https://arxiv.org/abs/2208.12937 Status source · Retrieved Sep 14, 2026 - Pseudodifferential arithmetic, Riemann and Lindelöf hypotheses withdrawn v48 https://arxiv.org/abs/2208.12937v48 Status source · Retrieved Sep 14, 2026 Problem familyThe Lindeloef Hypothesis on the Growth of the Riemann Zeta Function · 1 problem in this edition. Family membership alone does not assert an implication. - 82 · The Lindeloef Hypothesis on the Growth of the Riemann Zeta Function this problem Ranking uncertainty90% source-fit bootstrap rank range: 70–97. Source-fit opponent count: 29. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 82. Rank 083 Zilber-Pink Conjecture on Unlikely Intersections Must every irreducible complex subvariety of a mixed Shimura variety contain only finitely many maximal pieces where intersection with a special subvariety is larger than dimension counting predicts? These oversized pieces are the conjecture’s atypical intersections.Number theory & arithmetic geometry Exact statementLet S be a mixed Shimura variety and W an irreducible subvariety of S defined over C. Call a subvariety A of W atypical if there is a special subvariety T of S with A contained in the intersection of T and W and dim A dim W + dim T - dim S. Prove that the union of all atypical subvarieties of W is a finite union of atypical subvarieties, equivalently that W contains only finitely many maximal atypical subvarieties. Why it mattersA proof would organize all these unexpectedly large intersections into finitely many maximal exceptional pieces. This would give a finite geometric description even when the variety meets infinitely many special subvarieties. Status Open · subcases solved Zilber–Pink remains open, with solved subcases for the full mixed-Shimura target. Strong non-Qbar/Hodge-generic hypotheses and pure-Shimura cases do not cover it; current universal mixed-scope certification remains limited. Status reviewed Sources - J. Pila, 'Combining the conjectures of Schanuel and Zilber-Pink', Rend. Lincei Mat. Appl. 36 2025 , 653-670, DOI 10.4171/RLM/1086; accepted 1 July 2025. https://ems.press/content/serial-article-files/52570 Statement source - The Zilber-Pink conjecture for varieties not defined over Qbar https://arxiv.org/html/2504.00865v1 Status source · Retrieved Sep 14, 2026 - A note on unlikely intersections in Shimura varieties https://arxiv.org/html/2209.07967 Status source · Retrieved Sep 14, 2026 Problem familyZilber-Pink Conjecture on Unlikely Intersections · 1 problem in this edition. Family membership alone does not assert an implication. - 83 · Zilber-Pink Conjecture on Unlikely Intersections this problem Ranking uncertainty90% source-fit bootstrap rank range: 71–102. Source-fit opponent count: 43. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 83. Rank 084 Weight–Monodromy Conjecture Monodromy organizes a variety’s l-adic cohomology into layers. Are the Frobenius eigenvalues on each layer algebraic, with the absolute sizes predicted by its degree and layer index in every complex embedding, for every smooth proper variety over a nonarchimedean local field and every prime l different from the residue characteristic?Number theory & arithmetic geometry Exact statementLet K be ANY nonarchimedean local field with finite residue field F q of characteristic p, X ANY proper smooth variety over K, i =0, and l ANY prime different from p. Put V=H^i et X over Kbar,Qbar l , retaining the original K. On an open inertia subgroup the action is unipotent and defines nilpotent monodromy N:V- V -1 , with its Galois-equivariant Tate twist. Let M be the unique increasing monodromy filtration centred at zero: N M j is contained in M j-2 -1 , and N^a:gr^M a V - gr^M -a V -a is an isomorphism for every a =0. For EVERY integer j and ANY lift Phi of geometric Frobenius of F q, every eigenvalue alpha on gr^M j V is conjectured to be algebraic and have absolute value q^ i+j /2 under ALL complex embeddings. This is purity of weight i+j. Defining N on an open inertia subgroup does not replace K or q. Finite-extension proof reductions must change cardinality and Frobenius consistently. Extending l-adic coefficients does not change the target. The equivalent source convention compares zero-centred M j with W i+j ; weight-filtration existence is not an added unresolved assertion. No restriction to projective, semistable, geometrically connected or special complete-intersection varieties is imposed. l=p, crystalline, torsion and arbitrary rigid-analytic analogues are not substituted. Why it mattersA proof would force every such eigenvalue to have absolute value q^ i + j /2 under every complex embedding. It would link local monodromy to prescribed Frobenius weights without imposing projectivity or semistability. Status Open · subcases solved Weight–monodromy remains open, with solved subcases. Complete intersections in abelian varieties are a subcase, and 2026 formality work assumes weight–monodromy. Rigid-analytic counterexamples or l=p analogues are not the stated smooth proper algebraic l = p statement. Status reviewed Sources - Peter Scholze, Perfectoid spaces, arXiv1111.4914v1,banner21November2011/internal27November2024,51pages https://arxiv.org/pdf/1111.4914v1 Statement source - Formality for rigid-analytic spaces satisfying the weight-monodromy conjecture https://arxiv.org/abs/2607.14517 Status source · Retrieved Sep 14, 2026 - Perfectoid covers of abelian varieties and the weight-monodromy conjecture https://arxiv.org/abs/2303.05610 Status source · Retrieved Sep 14, 2026 - Semistable Lefschetz pencils https://page.mi.fu-berlin.de/esnault/preprints/helene/151 mw.pdf Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyWeight–Monodromy Conjecture · 1 problem in this edition. Family membership alone does not assert an implication. - 84 · Weight–Monodromy Conjecture this problem Ranking uncertainty90% source-fit bootstrap rank range: 73–95. Source-fit opponent count: 54. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 84. Rank 085 Sarnak's Mobius randomness conjecture for zero-entropy systems The Möbius function is zero on integers divisible by a prime square, and otherwise is +1 or −1 according to the parity of the number of prime factors. Does it have zero average correlation with every continuous measurement along every orbit of a compact metric dynamical system with zero topological entropy?Probability, ergodic theory & dynamics Exact statementProve that for every compact metric topological dynamical system X,T of zero topological entropy, every continuous f on X, and every x in X, the average N^ -1 sum {n<=N} mu n f T^n x tends to zero as N tends to infinity. Why it mattersA proof would exclude persistent correlation between the Möbius function and any observable orbit in this entire zero-entropy class. A counterexample would locate structured dynamical behavior with a surviving arithmetic correlation. Status Open Sarnak remains open for ordinary n-iterate Cesaro averages. Polynomial-iterate counterexamples and a special infinite-torus theorem do not resolve it. An author-site disproof assertion lacks a verified exact counterexample or independent acceptance. Status reviewed Sources - Q. Liu, J. Ma, and H. Wang, The Mobius Disjointness Conjecture on infinite-dimensional torus, Introduction 2026 . https://arxiv.org/abs/2603.11087 Statement source - Minimal zero entropy subshifts can be unrestricted along any sparse set https://www.cambridge.org/core/journals/ergodic-theory-and-dynamical-systems/article/minimal-zero-entropy-subshifts-can-be-unrestricted-along-any-sparse-set/1357B8C6E5361EC6439CEA52634BD7A0 Status source · Retrieved Sep 14, 2026 - The Moebius Disjointness Conjecture on infinite-dimensional torus https://arxiv.org/abs/2603.11087 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familySarnak's Mobius randomness conjecture for zero-entropy systems · 1 problem in this edition. Family membership alone does not assert an implication. - 85 · Sarnak's Mobius randomness conjecture for zero-entropy systems this problem Ranking uncertainty90% source-fit bootstrap rank range: 67–101. Source-fit opponent count: 24. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 85. Rank 086 Campana's Conjecture Characterizing Special Varieties by Potential Density of Rational Points For a smooth projective geometrically integral variety over a number field, is Campana's geometric condition of specialness exactly what allows its rational points to become Zariski dense after a finite extension of the field? Specialness is tested over the algebraic closure.Number theory & arithmetic geometry Campana specialness and rational points · 1 problem family-context-problem-campana-specialness-potential-density Exact statementFor every number field K and every smooth projective geometrically integral variety X/K, prove or disprove that X over an algebraic closure of K is special in Campana's sense if and only if X L is Zariski dense in X for some finite extension L/K. Here specialness has its usual projective meaning: for each integer 1<=p<=dim X and rank-one coherent subsheaf F of the sheaf of algebraic p-forms on X over the algebraic closure, the Kodaira dimension kappa F is less than p equivalently, the Campana core is a point . A proof must establish the biconditional for all such X; one counterexample to either implication refutes it. This target excludes nonproper integral-point, weakly-special, analytic and function-field variants. Why it mattersThis would characterize an arithmetic property—potential density of rational points—by geometry. A proof would also give potential density for K3 surfaces and the Bombieri–Lang non-density assertion for positive-dimensional varieties of general type. Status Open · subcases solved Status reviewed Sources - Abramovich, Birational geometry for number theorists https://arxiv.org/pdf/math/0701105 Statement source Problem familyCampana specialness and rational points · 1 problem in this edition. Family membership alone does not assert an implication. - 86 · Campana's Conjecture Characterizing Special Varieties by Potential Density of Rational Points this problem Ranking uncertainty90% source-fit bootstrap rank range: 70–99. Source-fit opponent count: 27. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 86. Rank 087 Bose-Einstein Condensation in Interacting Continuous Bose Gases Can Bose–Einstein condensation be proved for an interacting gas of bosons moving in continuous three-dimensional space at low temperatures? The target is persistent off-diagonal long-range order as particle number and volume grow together at fixed density.Mathematical physics Exact statementProve mathematically that an interacting continuous Bose gas in three dimensions exhibits non-zero off-diagonal long-range order Bose-Einstein condensation in the thermodynamic limit at low temperatures. Why it mattersA proof would establish Bose–Einstein condensation in the stated interacting continuum setting. It would show that the relevant quantum correlations persist over arbitrarily large distances in the thermodynamic limit. Status Open Interacting continuum BEC remains provisionally open with an explicit claim/scope warning. Suto's 2023 high-density condensation claim potentially overlaps the existential wording, but acceptance and equivalence to fixed-density low-temperature thermodynamic ODLRO were not established. The independent April 2026 manuscript still calls interacting ODLRO open; interaction class and scaling need expert reconciliation. Status reviewed Sources - E. H. Lieb, R. Seiringer, J. P. Solovej, J. Yngvason, The Mathematics of the Bose Gas, Birkhäuser, 2005. https://doi.org/10.1007/3-7643-7338-9 Statement source - Bose-Einstein condensation of interacting bosons: A two-step proof https://arxiv.org/abs/2305.18959 Status source · Retrieved Sep 14, 2026 - The free energy of the interacting Bose gas: A variational description with loops and interlacements https://www.wias-berlin.de/people/koenig/www/CondensateInterlace.pdf Status source · Retrieved Sep 14, 2026 Problem familyBose-Einstein Condensation in Interacting Continuous Bose Gases · 1 problem in this edition. Family membership alone does not assert an implication. - 87 · Bose-Einstein Condensation in Interacting Continuous Bose Gases this problem Ranking uncertainty90% source-fit bootstrap rank range: 74–104. Source-fit opponent count: 26. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 87. Rank 088 Elliott-Halberstam Conjecture on the Level of Distribution of Primes When counting primes in arithmetic progressions, can their distribution errors be controlled for step sizes nearly as large as the counting limit x? Elliott–Halberstam predicts arbitrarily strong logarithmic savings in the sum of the largest errors over residue classes and shorter counting ranges, for step sizes up to x^θ with any 0 < θ < 1.Number theory & arithmetic geometry Exact statementProve that for every A 0 and every 0 < theta < 1, the sum over q <= x^theta of the maximal Chebyshev error |psi y; q, a - y/phi q | over y <= x and a,q =1 is O x / log x ^A , extending Bombieri-Vinogradov from theta < 1/2 to every theta < 1. Why it mattersA proof would control the combined error in prime distribution across arithmetic progressions for moduli extending to every fixed power below x. The estimate controls the sum of worst errors for each modulus, retaining uniformity over shorter cutoffs. Status Open Elliott–Halberstam remains open unconditionally. A recent implication assumes both GRH and Dirichlet-L pair correlation, while restricted-modulus results do not prove the stated all-moduli maximal-error estimate. Status reviewed Sources - Weisstein, E. W., Elliott-Halberstam Conjecture, From MathWorld - A Wolfram Resource, updated 16 August 2026. https://mathworld.wolfram.com/Elliott-HalberstamConjecture.html Statement source - Pair Correlation of zeros of Dirichlet L-Functions: A possible path towards the conjectures of Chowla, Elliott-Halberstam and Montgomery https://arxiv.org/abs/2411.19762 Status source · Retrieved Sep 14, 2026 - Primes in arithmetic progressions to large moduli III: Uniform residue classes https://arxiv.org/html/2006.08250 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyElliott-Halberstam Conjecture on the Level of Distribution of Primes · 1 problem in this edition. Family membership alone does not assert an implication. - 88 · Elliott-Halberstam Conjecture on the Level of Distribution of Primes this problem Ranking uncertainty90% source-fit bootstrap rank range: 76–108. Source-fit opponent count: 31. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 88. Rank 089 Grothendieck–Teichmüller Conjecture The absolute Galois group of the rationals describes the field symmetries of algebraic numbers. Does its standard Ihara embedding account for every element of Drinfeld’s original profinite Grothendieck–Teichmüller group, which encodes compatible symmetries of fundamental-group data?Number theory & arithmetic geometry Grothendieck–Teichmüller theory · 2 problems family-context-problem-grothendieck-teichmuller-conjecture Exact statementLet G Q=Gal Qbar/Q . Let GT-hat be Drinfeld's original profinite Grothendieck-Teichmuller group, including its standard commutator and invertibility conditions, not merely the pair-equation monoid, gentle GT, a pro-l or prounipotent version, or a higher-genus subgroup. Fix the standard Ihara embedding Ih:G Q - GT-hat specified by the Ihara splitting in Bortnovskyi-Dolgushev-Holikov-Pashkovskyi Section1.3, equations1.5-1.6: Ih g = chi g -1 /2,f g in that source's coordinates, with chi the cyclotomic character, and associated action x - x^chi g , y - f g^ -1 y^chi g f g on the profinite free group. The conjecture asserts that this actual continuous injective homomorphism is surjective, hence an isomorphism of profinite groups. The source's group law and fixed splitting are retained; formulas from an opposite composition convention are not mixed into this definition. Why it mattersSurjectivity would identify that entire profinite group with the arithmetic Galois group through the actual embedding. Failure would exhibit elements satisfying the group’s defining conditions that do not arise from Galois symmetries. Status Open Grothendieck–Teichmuller remains open for surjectivity of the fixed Ihara embedding into the original full profinite GT. August 2026 quotient results and underlying-space homeomorphisms do not prove this group-homomorphism surjectivity. Status reviewed Sources - Ivan Bortnovskyi, Vasily A. Dolgushev, Borys Holikov, Vadym Pashkovskyi, Accessing non-abelian quotients of the Grothendieck-Teichmueller group via elementary tools, arXiv2405.11725v3,6August2026,35pages https://arxiv.org/pdf/2405.11725v3 Statement source - Accessing non-abelian quotients of the Grothendieck-Teichmueller group via elementary tools https://arxiv.org/html/2405.11725 Status source · Retrieved Sep 14, 2026 - Galois actions on surfaces and a higher genus Grothendieck-Teichmüller group https://arxiv.org/html/2606.01466v1 Status source · Retrieved Sep 14, 2026 - Proving the Grothendieck–Teichmüller Conjecture for Profinite Spaces & The Galois Grothendieck Path Integral https://arxiv.org/html/2503.13006 Status source · Retrieved Sep 14, 2026 Problem familyGrothendieck–Teichmüller theory · 2 problems in this edition. Family membership alone does not assert an implication. - 89 · Grothendieck–Teichmüller Conjecture this problem - 449 · Deligne–Drinfeld Freeness Conjecture for grt 1 problem-top500-omission-w045049-deligne-drinfeld-grt-one-freeness Ranking uncertainty90% source-fit bootstrap rank range: 79–102. Source-fit opponent count: 64. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 89. Rank 090 Deligne's Absolute Hodge Conjecture Hodge classes are cohomology classes singled out by a variety's complex geometry. On every smooth projective complex variety, must each rational Hodge class remain one after any automorphism of the complex number field?Number theory & arithmetic geometry Hodge conjectures · 3 problems family-context-problem-top500-omission-w045049-absolute-hodge-conjecture Exact statementFor every smooth projective variety X over C, every p =0 and every rational Hodge class alpha in H^ 2p X,Q p , alpha is absolute Hodge: for every automorphism sigma of C, its sigma-conjugate under algebraic de Rham comparison on X^sigma is again a rational Hodge class, with the same Tate-twist convention. Why it mattersA proof would make the Hodge property invariant under all these field automorphisms, using the specified de Rham comparison and Tate twist. A counterexample would identify a class whose conjugate loses that property. Status Open · subcases solved Status review date not recorded in this edition Sources Problem familyHodge conjectures · 3 problems in this edition. Family membership alone does not assert an implication. - 4 · Hodge Conjecture problem-hodge-conjecture - 24 · Grothendiecks Generalized Hodge Conjecture problem-grothendiecks-generalized-hodge-conjecture - 90 · Deligne's Absolute Hodge Conjecture this problem Ranking uncertainty90% source-fit bootstrap rank range: 78–104. Source-fit opponent count: 39. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 90. Rank 091 The Mumford-Tate Conjecture for Abelian Varieties over Number Fields An abelian variety has symmetries seen through both number theory and complex geometry. Over any number field, do the Lie algebras of its l-adic Galois image and its Mumford–Tate group agree after passing to l-adic coefficients, so that the two descriptions give the same infinitesimal symmetries?Number theory & arithmetic geometry Exact statementProve that for every abelian variety A over a number field K, the Lie algebra of the l-adic Galois image Lie Gal Kbar/K equals the Lie algebra of the Mumford-Tate group MT A tensor Q l. Why it mattersA proof would identify the specified arithmetic Galois symmetries with the Mumford–Tate description at the Lie-algebra level. A counterexample would expose a mismatch between these two symmetry measurements for an abelian variety. Status Open Mumford–Tate remains open for every number-field abelian variety. Frobenius-field progress does not prove equality of the two Lie algebras universally; a rendered 2026 HTML date is not an authenticated revision of the 2024 source. Status reviewed Sources - J.-P. Serre, Abelian l-Adic Representations and Elliptic Curves, Addison-Wesley, 1968. https://doi.org/10.1201/9781439864227 Statement source - On the Frobenius fields of abelian varieties over number fields https://arxiv.org/html/2402.07935v2 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyThe Mumford-Tate Conjecture for Abelian Varieties over Number Fields · 1 problem in this edition. Family membership alone does not assert an implication. - 91 · The Mumford-Tate Conjecture for Abelian Varieties over Number Fields this problem Ranking uncertainty90% source-fit bootstrap rank range: 81–106. Source-fit opponent count: 35. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 91. Rank 092 Entanglement Area Law for 2D Gapped Quantum Systems For locally interacting quantum spins on a square lattice, with a unique ground state and a size-independent positive energy gap, must every region’s entanglement entropy be at most a constant times the number of edges crossing its boundary? The constant may depend on the fixed local parameters and gap, but not on the region or lattice size.Quantum information & computation Exact statementProve the two-dimensional ground-state area-law conjecture for finite-dimensional lattice spins: for families of finite 2D square-lattice Hamiltonians with fixed on-site dimension, uniformly bounded finite-range local interactions, a unique ground state, and a positive spectral-gap lower bound independent of system size, the ground-state entanglement entropy S rho A is at most C times the boundary size |partial A| for every region A. The constant C may depend on the fixed local parameters and gap lower bound, but not on the system size or region. Here rho A is the reduced pure-ground-state density operator and boundary size counts lattice edges crossing from A to its complement. Why it mattersA proof would limit entanglement by boundary size throughout this class, including frustrated Hamiltonians. The constant could depend on fixed interaction parameters and the gap, while remaining valid as the system and chosen region grow. Status Open The universal two-dimensional gapped ground-state area law remains open. The 2026 AGSP theorem adds conditions available in restricted frustration-free locally gapped cases, not every globally gapped unique-ground-state Hamiltonian. Status reviewed Sources - I. Arad, R. Firanko, R. Jain, Area Laws and Tensor Networks for Maximally Mixed Ground States, Communications in Mathematical Physics 407, 54 2026 . https://link.springer.com/article/10.1007/s00220-026-05554-z Statement source - Area Laws and Tensor Networks for Maximally Mixed Ground States https://link.springer.com/article/10.1007/s00220-026-05554-z Status source · Retrieved Sep 14, 2026 - Universal lower bound on topological entanglement entropy https://arxiv.org/abs/2302.00689 Status source · Retrieved Sep 14, 2026 Problem familyEntanglement Area Law for 2D Gapped Quantum Systems · 1 problem in this edition. Family membership alone does not assert an implication. - 92 · Entanglement Area Law for 2D Gapped Quantum Systems this problem Ranking uncertainty90% source-fit bootstrap rank range: 79–107. Source-fit opponent count: 24. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 92. Rank 093 Grothendieck Standard Conjecture of Hodge Type After identifying algebraic cycles with the same intersection numbers, does the sign-corrected intersection form give a positive value on every nonzero primitive class? This is asked in codimensions up to half the dimension, for every ample class on smooth projective irreducible varieties over algebraically closed fields of any characteristic.Algebra, representation & category theory Grothendieck’s standard conjectures · 2 problems family-context-problem-grothendieck-standard-conjecture-hodge-type Exact statementFor every smooth projective irreducible variety X of dimension n over an algebraically closed field k of any characteristic and every ample divisor class L, put N^i X Q equal to codimension-i algebraic cycles with rational coefficients modulo numerical equivalence, with N^j=0 outside 0<=j<=n. For each integer 0<=i<=n/2 let P L^i=ker L^ n-2i+1 :N^i X Q - N^ n-i+1 X Q . The numerical standard conjecture of Hodge type asserts that q L^i a,b = -1 ^i deg a.b.L^ n-2i is positive definite on P L^i after extending scalars to R. This is numerical primitive positivity, not an unconditional identification with primitive algebraic cohomology for an arbitrary Weil theory; it does not assume or require proving the other standard conjectures separately. Why it mattersA proof would give a positive inner-product structure to these primitive numerical cycle spaces. The conclusion concerns cycles modulo numerical equivalence and does not itself identify them with primitive classes in an arbitrary cohomology theory. Status Open · subcases solved Standard Conjecture of Hodge type remains open, with solved subcases. Fourfolds, powers of threefolds and other specified examples do not establish numerical primitive positivity in every dimension and characteristic; newer preprint proofs were not independently certified. Status reviewed Sources - Thomas Agugliaro, Examples for the standard conjecture of Hodge type, Doc.Math.31 2026 855-904, DOI10.4171/DM/1057 https://ems.press/content/serial-article-files/53071 Statement source - Examples for the standard conjecture of Hodge type https://ems.press/content/serial-article-files/53071 Status source · Retrieved Sep 14, 2026 - Standard conjectures for abelian fourfolds https://link.springer.com/article/10.1007/s00222-020-00990-7 Status source · Retrieved Sep 14, 2026 - Standard conjecture of Hodge type for powers of abelian varieties https://arxiv.org/abs/2510.21562 Status source · Retrieved Sep 14, 2026 Problem familyGrothendieck’s standard conjectures · 2 problems in this edition. Family membership alone does not assert an implication. - 43 · Grothendieck's Standard Conjecture D problem-grothendieck-standard-conjecture-d - 93 · Grothendieck Standard Conjecture of Hodge Type this problem Ranking uncertainty90% source-fit bootstrap rank range: 79–110. Source-fit opponent count: 49. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 93. Rank 094 PSPACE versus EXPTIME If a deterministic algorithm solves a yes-or-no problem in time 2^ n^k for some fixed k, must another deterministic algorithm solve it using only polynomially many working-memory bits in the input length n?Theoretical computer science Exact statementDetermine whether PSPACE=EXPTIME for languages over finite binary strings in the standard deterministic multitape Turing-machine model with a read-only input tape and work tapes. PSPACE is the union over integers k =1 of DSPACE n^k , with asymptotic constant factors allowed and read-only input cells not counted as work space. EXPTIME is the union over integers k =1 of DTIME 2^ n^k , with asymptotic constant factors allowed. Machines are total deciders that halt on every input. This is an unrelativized equality of the full classes, not of one fixed resource exponent. Why it mattersAn equality would show that polynomial memory suffices for every computation in the full exponential-time class. A separation would identify a decision problem whose time bound can be exponential but whose memory requirements exceed every polynomial. Status Open PSPACE versus EXPTIME remains open with bounded recency evidence. The Czerwinski separation was withdrawn in March 2024 with an independent critique, which defeats that claim but does not certify absence of every later one. Status reviewed Sources - Alberto Pettorossi, Elements of Computability, Decidability, and Complexity, Third Edition, printed115, item iii ; Arora–Barak Chapter4 machine conventions. https://art.torvergata.it/bitstream/2108/32769/60437/ComputabilityDecidabilityComplexity.pdf Statement source - Separation of PSPACE and EXP withdrawn https://arxiv.org/abs/2104.14316 Status source · Retrieved Sep 14, 2026 - A Critique of Czerwinski's Separation of PSPACE and EXP https://arxiv.org/abs/2304.14575 Status source · Retrieved Sep 14, 2026 Problem familyPSPACE versus EXPTIME · 1 problem in this edition. Family membership alone does not assert an implication. - 94 · PSPACE versus EXPTIME this problem Ranking uncertainty90% source-fit bootstrap rank range: 80–108. Source-fit opponent count: 57. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 94. Rank 095 Kannan-Lovasz-Simonovits conjecture on isoperimetry of log-concave measures For a log-concave probability distribution, can a hyperplane cut achieve a boundary-to-mass ratio within a universal constant factor of the best possible cut, in every dimension? Each cut’s boundary is normalized by the smaller of the two masses it separates.Applied & computational mathematicsResolution claim under review · historical rank Exact statementProve that there is a universal constant C 0 such that, for every log-concave probability measure mu on R^n, its Cheeger isoperimetric constant is at least C times the infimum of mu^+ H /min mu H , 1-mu H over halfspaces H. Equivalently, prove the dimension-free KLS lower bound psi mu = C/sqrt ||Cov mu || op , with a universal constant C. Why it mattersA proof would make simple halfspace cuts sufficient, up to a universal factor, to describe the measure’s worst isoperimetric bottleneck. The bound would remain independent of dimension and admit the stated covariance formulation. Status Resolution claim under review Claimed pending refereeing — full KLS proof claims. Song and Zhang v2 and Bizeul, Klartag and Lehec v1 claim a universal Poincaré bound for isotropic log-concave measures, giving the dimension-free Cheeger target stated here. Song–Zhang v1 only gave an unbounded log-star bound. Bizeul–Klartag–Lehec build on the earlier Song–Zhang criterion, so these are not wholly independent confirmations. The pinned October 6 manuscript by Balasubramanian and Kasiviswanathan makes another full claim. No public Lean formalization or kernel receipt was located in the reviewed sources; this was a source-and-scope review, not a complete proof audit. Status reviewed Sources - Afonso S. Bandeira, Daniil Dmitriev, Kevin Lucca, Petar Nizić-Nikolac, and Almut Rödder, "Randomstrasse101: Open Problems of 2025," arXiv:2603.29571, 2026. https://arxiv.org/pdf/2603.29571 Statement source - Song and Zhang, v1: the earlier dimension-dependent log-star bound https://arxiv.org/abs/2610.01447v1 Status source · Retrieved Oct 6, 2026 - Song and Zhang, v2: full dimension-free KLS claim Theorems 1.1 and 9.1 https://arxiv.org/html/2610.01447v2 Status source · Retrieved Oct 6, 2026 - Bizeul, Klartag and Lehec, v1: full KLS claim Theorem 1.1 https://arxiv.org/html/2610.05474v1 Status source · Retrieved Oct 6, 2026 - Balasubramanian and Kasiviswanathan: full KLS claim, pinned October 6 manuscript https://raw.githubusercontent.com/kriznakumar/paper/4837c33649ba2271f43c9684e9350ecbdd725f95/KLS.pdf Status source · Retrieved Oct 6, 2026 Research on ProofAtlas Problem familyKannan-Lovasz-Simonovits conjecture on isoperimetry of log-concave measures · 1 problem in this edition. Family membership alone does not assert an implication. - 95 · Kannan-Lovasz-Simonovits conjecture on isoperimetry of log-concave measures this problem Ranking uncertainty90% source-fit bootstrap rank range: 81–115. Source-fit opponent count: 33. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 95. Rank 096 Effective Mordell Conjecture for Curves of Genus at Least Two Can we give a guaranteed terminating algorithm listing every point defined over a number field on any smooth projective curve of genus at least two, or an explicit bound on the points’ arithmetic size, called height, that makes such a search finite?Number theory & arithmetic geometry Exact statementFor every smooth projective curve X of genus at least two over a number field K, give an explicit height bound for X K , or an algorithm guaranteed to terminate that computes all points of X K . Why it mattersA resolution would make the curve’s rational points accessible through a provably complete search. An explicit height bound or guaranteed terminating enumeration would provide a stopping criterion, so a list could be known to omit no points. Status Open · subcases solved Effective Mordell remains open, with solved subcases for one uniform algorithm taking both field and curve as input. The 2025 automorphism/rank-dependent height theorem is restricted, not an all-curves terminating enumeration. Status reviewed Sources - Natalia Garcia-Fritz and Hector Pasten, "Effective Mordell for curves with enough automorphisms", arXiv:2503.10443 \ math.NT\ , v1 dated 13 March 2025. https://arxiv.org/pdf/2503.10443 Statement source - Effective Mordell for curves with enough automorphisms https://arxiv.org/abs/2503.10443 Status source · Retrieved Sep 14, 2026 - Computing genus 2 curves over Q whose Jacobian has good reduction away from 2 https://warwick.ac.uk/fac/sci/maths/people/staff/visser/bmc2024 slides.pdf Status source · Retrieved Sep 14, 2026 Problem familyEffective Mordell Conjecture for Curves of Genus at Least Two · 1 problem in this edition. Family membership alone does not assert an implication. - 96 · Effective Mordell Conjecture for Curves of Genus at Least Two this problem Ranking uncertainty90% source-fit bootstrap rank range: 80–116. Source-fit opponent count: 25. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 96. Rank 097 L versus NL Can a deterministic algorithm decide whether a directed graph has a path from one chosen vertex to another using only logarithmically many bits of working memory relative to input length? This is equivalent to asking whether nondeterminism adds power when working memory is logarithmic.Theoretical computer science Exact statementDetermine whether every language decidable by a nondeterministic Turing machine using O log n work space is decidable by a deterministic Turing machine using O log n work space; equivalently, decide whether directed s-t reachability belongs to L. Why it mattersEquality would remove nondeterminism without increasing the logarithmic workspace bound, yielding such an algorithm for directed reachability. Separation would show that nondeterministic choices increase computational power at this memory scale. Status Open Status review date not recorded in this edition Sources Research on ProofAtlas Problem familyL versus NL · 1 problem in this edition. Family membership alone does not assert an implication. - 97 · L versus NL this problem Ranking uncertainty90% source-fit bootstrap rank range: 82–116. Source-fit opponent count: 33. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 97. Rank 098 Littlewood Conjecture For every pair of real numbers α and β, can q times the product of the distances from qα and qβ to their nearest integers become arbitrarily small for arbitrarily large positive integers q?Probability, ergodic theory & dynamics Exact statementProve or disprove that for every pair of real numbers alpha and beta, liminf {q to infinity} q ||q alpha|| ||q beta|| = 0. Why it mattersA proof would guarantee this multiplicative simultaneous approximation for every real pair. The product allows an exceptionally close approximation to one coordinate to compensate for a less close approximation to the other. Status Open Classical Littlewood remains open. The March 2026 disproof and August positive-dimensional counterexamples concern the different uniform Littlewood conjecture, not the stated real-pair liminf assertion. Status reviewed Sources - Steven Robertson, Combinatorics on Number Walls and the P t -adic Littlewood Conjecture, Section 1, 2025 revision. https://arxiv.org/html/2307.00955v3 Statement source - Disproof of the uniform Littlewood conjecture https://arxiv.org/abs/2603.12611 Status source · Retrieved Sep 14, 2026 - The uniform Littlewood conjecture fails on a set of positive Hausdorff dimension https://arxiv.org/abs/2608.25059 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyLittlewood Conjecture · 1 problem in this edition. Family membership alone does not assert an implication. - 98 · Littlewood Conjecture this problem Ranking uncertainty90% source-fit bootstrap rank range: 71–129. Source-fit opponent count: 24. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 98. Rank 099 Grothendiecks Section Conjecture in Anabelian Geometry Does the arithmetic fundamental group, which combines geometric loops with field symmetries, encode the rational points of every smooth complete curve of genus at least two over a number field exactly as ways to split its projection to field symmetries, counted up to conjugacy?Number theory & arithmetic geometry Exact statementProve that for every smooth, complete hyperbolic curve C of genus g = 2 over a number field K, the map sending K-rational points C K to conjugacy classes of splittings of the core exact sequence 1 - pi 1 C {Kbar} - pi 1 C - Gal Kbar/K - 1 is a bijection. Why it mattersA proof would recover exactly the rational points from group-theoretic splittings. Every splitting class would come from one rational point, turning this fundamental-group structure into a precise description of the curve’s arithmetic points. Status Open The number-field section conjecture remains open with explicit current-status uncertainty. Recent conditional finite-type consequences and restricted affine S-Selmer sections do not establish all sections for every smooth complete hyperbolic number-field curve. Status reviewed Sources - A. Grothendieck, Brief an G. Faltings, LMS Lecture Notes 242:49-58, 1997. https://doi.org/10.1017/CBO9780511758874.005 Statement source - Polylogarithmic Chabauty--Kim loci over number fields https://arxiv.org/abs/2608.20615 Status source · Retrieved Sep 14, 2026 - On the section conjecture over fields of finite type https://onlinelibrary.wiley.com/doi/full/10.1002/mana.70049 Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyGrothendiecks Section Conjecture in Anabelian Geometry · 1 problem in this edition. Family membership alone does not assert an implication. - 99 · Grothendiecks Section Conjecture in Anabelian Geometry this problem Ranking uncertainty90% source-fit bootstrap rank range: 78–125. Source-fit opponent count: 28. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 99. Rank 100 Green-Griffiths-Lang Conjecture on entire curves in varieties of general type An entire curve is a holomorphic map from the whole complex plane. On a projective variety of general type, must each such curve stay inside some proper algebraic subvariety?Geometry & topologyReviewed hold · conditional rank Exact statementProve the Green-Griffiths-Lang conjecture: for a projective variety X of general type, every entire holomorphic curve f: C - X factors through a proper closed subvariety equivalently, jet differentials define an algebraic differential equation for non-constant entire curves . Why it mattersA proof would prevent any entire curve from spreading densely through a general-type variety in the Zariski sense. It would connect the variety’s algebraic type to a restriction on globally defined complex-analytic maps. Status Reviewed hold · conditional rank Recorded hold reason. Green–Griffiths–Lang remains open with its inherited scope warning. The standard uniform exceptional locus and the frozen per-curve wording must be distinguished; the asserted equivalence with jet-differential equations is not certified. High-degree generic hypersurfaces and plane-curve complements do not resolve all projective general-type varieties.Green–Griffiths–Lang remains open with its inherited scope warning. The standard uniform exceptional locus and the stated per-curve wording must be distinguished; the asserted equivalence with jet-differential equations is not certified. High-degree generic hypersurfaces and plane-curve complements do not resolve all projective general-type varieties. Status reviewed Sources - J.-P. Demailly, Holomorphic Morse inequalities and the Green-Griffiths-Lang conjecture, arXiv:1011.3636 \ math.AG\ , 2010. https://arxiv.org/abs/1011.3636 Statement source - Computing Jet Differentials and the Green-Griffiths-Lang Conjecture for Complements of Smooth Plane Curves https://arxiv.org/abs/2608.20479 Status source · Retrieved Sep 14, 2026 - Non-reductive geometric invariant theory and hyperbolicity https://link.springer.com/article/10.1007/s00222-023-01219-z Status source · Retrieved Sep 14, 2026 Research on ProofAtlas Problem familyGreen-Griffiths-Lang Conjecture on entire curves in varieties of general type · 1 problem in this edition. Family membership alone does not assert an implication. - 100 · Green-Griffiths-Lang Conjecture on entire curves in varieties of general type this problem Ranking uncertainty90% source-fit bootstrap rank range: 81–118. Source-fit opponent count: 30. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 100. Rank 101 Classical Szpiro Discriminant–Conductor Conjecture over Q Can the magnitude of an elliptic curve's minimal discriminant be controlled by its conductor—two measures of its behavior at bad primes—using every power greater than six, with a bound uniform over all curves defined over the rationals?Number theory & arithmetic geometry abc and Szpiro conjectures · 2 problems family-context-problem-top500-omission-w050054-classical-szpiro-discriminantconductor-conjecture-over-q Exact statementProve or disprove that for every epsilon 0 there is C epsilon 0 such that every elliptic curve E over Q with minimal discriminant Delta E and conductor N E satisfies |Delta E| <= C epsilon N E^ 6+epsilon . Why it mattersA proof would uniformly control the discriminant in terms of the conductor, with the constant depending only on ε. Failure would mean that some positive ε admits no such uniform constant. Status Open · subcases solved Status review date not recorded in this edition Sources Problem familyabc and Szpiro conjectures · 2 problems in this edition. Family membership alone does not assert an implication. - 9 · The abc Conjecture of Masser and Oesterle problem-the-abc-conjecture-of-masser-and-oesterle - 101 · Classical Szpiro Discriminant–Conductor Conjecture over Q this problem Ranking uncertainty90% source-fit bootstrap rank range: 82–119. Source-fit opponent count: 25. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 101. Rank 102 Bunyakovsky conjecture If a polynomial with integer coefficients cannot split into lower-degree integer polynomials, eventually takes positive values, and has no prime dividing every value, must it produce primes at infinitely many positive integer inputs?Number theory & arithmetic geometryOpen · disputed claim Exact statementProve or disprove that every irreducible polynomial f in Z x with positive leading coefficient and no fixed prime divisor takes prime values for infinitely many positive integer inputs. Why it mattersA proof would establish infinitely many prime values for every polynomial satisfying these conditions, including x² + 1. A counterexample would reveal an obstruction beyond factorization, sign and fixed prime divisors. Status Open · disputed claim Status review date not recorded in this edition Sources Problem familyBunyakovsky conjecture · 1 problem in this edition. Family membership alone does not assert an implication. - 102 · Bunyakovsky conjecture this problem Ranking uncertainty90% source-fit bootstrap rank range: 77–134. Source-fit opponent count: 25. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 102. Rank 103 Superpolynomial Lower Bounds for Frege Propositional Proof Systems In standard Frege systems for reasoning with Boolean formulas, can we exhibit an explicit family of always-true formulas whose proofs exceed every polynomial bound in the formula length, measured by total proof size or by the number of proof lines?Theoretical computer science Classical complexity separations · 5 problems family-context-problem-superpolynomial-lower-bounds-for-frege-propositional-proof-systems Exact statementProve a superpolynomial lower bound on the size, or number of lines, of proofs in standard Frege propositional proof systems for an explicit family of propositional tautologies. Why it mattersA proof would give concrete logical truths for which every proof in these systems must become very large. It would establish an intrinsic limitation of the proof system, beyond difficulties in choosing a proof-search method. Status Open Status review date not recorded in this edition Sources - J. Krajicek, Proof Complexity, Cambridge University Press, 2019. https://doi.org/10.1017/9781108643818 Statement source Problem familyClassical complexity separations · 5 problems in this edition. Family membership alone does not assert an implication. - 1 · P versus NP problem-p-versus-np - 13 · NP versus coNP Problem problem-np-versus-conp-problem - 14 · P versus PSPACE problem-p-versus-pspace - 34 · P versus NP ∩ coNP problem-top500-omission-w050054-p-versus-np-conp - 103 · Superpolynomial Lower Bounds for Frege Propositional Proof Systems this problem Ranking uncertainty90% source-fit bootstrap rank range: 88–125. Source-fit opponent count: 35. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 103. Rank 104 Two-Dimensional Jacobian Conjecture in Characteristic Zero If a polynomial transformation of a plane has a nonzero constant determinant of derivatives, must it have an inverse that is also polynomial, over every field of characteristic zero?Algebra, representation & category theory Exact statementLet k be a field of characteristic zero and let F:k^2 - k^2 be a polynomial map whose Jacobian determinant is a nonzero constant. Determine whether F must have a polynomial inverse, equivalently whether every two-variable Keller map is a polynomial automorphism of the affine plane. Why it mattersA proof would turn this differential condition into global polynomial reversibility for every planar map in scope. A counterexample would show that constant nonzero Jacobian determinant does not guarantee an algebraic inverse. Status Open Gao's July 31, 2026 preprint reports counterexamples only in dimensions greater than two and explicitly leaves the two-dimensional case open. This higher-dimensional claim does not resolve this entry. October 6, 2026: Van den Essen explains the three-variable Jacobian counterexample. The current ranked entity explicitly concerns two variables. The complete proof was not audited in this review. Status reviewed Sources - Terence Tao, "A digestion of the Jacobian conjecture counterexample", What's new, 21 July 2026. https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/ Statement source - Shuhong Gao, Counterexamples to the Jacobian conjecture in dimensions greater than two https://arxiv.org/abs/2608.00222 Status source · Retrieved Sep 16, 2026 - Gao, section 2: Background and previous work https://arxiv.org/html/2608.00222v1 S2 Status source · Retrieved Sep 16, 2026 - Two-Dimensional Jacobian Conjecture in Characteristic Zero: October 2026 scope review https://arxiv.org/abs/2609.17795 Status source · Retrieved Oct 6, 2026 Problem familyTwo-Dimensional Jacobian Conjecture in Characteristic Zero · 1 problem in this edition. Family membership alone does not assert an implication. - 104 · Two-Dimensional Jacobian Conjecture in Characteristic Zero this problem Ranking uncertainty90% source-fit bootstrap rank range: 83–129. Source-fit opponent count: 25. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 104. Rank 105 Polignac's conjecture Does every positive even number occur infinitely often as the gap between two consecutive primes?Number theory & arithmetic geometry Exact statementFor every positive even integer 2k, there are infinitely many pairs of consecutive primes whose difference is 2k. Why it mattersA proof would show that every positive even gap recurs indefinitely among neighboring primes. A disproof would identify an even gap that eventually disappears or never occurs in consecutive-prime spacings. Status Open Status review date not recorded in this edition Sources Problem familyPolignac's conjecture · 1 problem in this edition. Family membership alone does not assert an implication. - 105 · Polignac's conjecture this problem Ranking uncertainty90% source-fit bootstrap rank range: 86–132. Source-fit opponent count: 24. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 105. Rank 106 Selberg 1/4 Conjecture On every congruence quotient of the upper half-plane, an arithmetic hyperbolic space, is the first nonzero Laplacian eigenvalue at least one-quarter? These eigenvalues describe how functions oscillate on the space.Number theory & arithmetic geometry Exact statementProve that the first nonzero eigenvalue lambda 1 of the Laplace-Beltrami operator on every congruence quotient of the upper half-plane is at least 1/4. Why it mattersA proof would exclude all nonzero Laplace eigenvalues below one-quarter throughout this arithmetic family of hyperbolic surfaces. It would provide one uniform spectral gap despite the different geometries of the quotients. Status Open Selberg 1/4 remains open with recency and claim-verification limits. Random covers, high-energy statistics and small-level checks do not cover every congruence quotient; the private formal-route assertion lacks an independently verified exact-target transfer. Status reviewed Sources - Formal statement, Chapter 4, Selberg property https://www.ma.huji.ac.il/~alexlub/BOOKS/On%20property/On%20property.pdf Statement source - Asymptotic independence for random permutations from surface groups https://link.springer.com/article/10.1007/s10711-025-01003-8 Status source · Retrieved Sep 14, 2026 - Twist-minimal trace formulas and the Selberg eigenvalue conjecture https://arxiv.org/abs/1803.06016 Status source · Retrieved Sep 14, 2026 Problem familySelberg 1/4 Conjecture · 1 problem in this edition. Family membership alone does not assert an implication. - 106 · Selberg 1/4 Conjecture this problem Ranking uncertainty90% source-fit bootstrap rank range: 87–140. Source-fit opponent count: 24. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 106. Rank 107 Global Regularity of the 3D Incompressible Magnetohydrodynamic Equations For an incompressible conducting fluid filling three-dimensional space, do all sufficiently smooth divergence-free velocity and magnetic fields evolve uniquely and smoothly forever when both viscosity and magnetic diffusion are positive?Analysis & PDE Incompressible fluid regularity · 2 problems family-context-problem-global-regularity-of-the-3d-incompressible-magnetohydrodynamic-equations Exact statementProve or disprove that every sufficiently smooth divergence-free initial velocity and magnetic field on R^3 generates a unique global smooth solution of the standard incompressible three-dimensional MHD system with positive viscosity and magnetic diffusivity. Why it mattersA proof would ensure that the coupled fluid and magnetic equations preserve smooth evolution for every datum in the stated class. A disproof would reveal a breakdown of regularity or uniqueness despite both forms of dissipation. Status Open · subcases solved Status review date not recorded in this edition Sources Problem familyIncompressible fluid regularity · 2 problems in this edition. Family membership alone does not assert an implication. - 6 · Navier-Stokes Existence and Smoothness: Fefferman A problem-navier-stokes-existence-and-smoothness - 107 · Global Regularity of the 3D Incompressible Magnetohydrodynamic Equations this problem Ranking uncertainty90% source-fit bootstrap rank range: 82–135. Source-fit opponent count: 24. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 107. Rank 108 Invariant Subspace Problem for Hilbert Space Operators Must every bounded linear operator on a separable infinite-dimensional complex Hilbert space map some closed linear subspace into itself, other than the zero subspace or the whole space?Analysis & PDE Exact statementDoes every bounded linear operator T on an infinite-dimensional separable complex Hilbert space H have a nontrivial closed invariant subspace, that is, a closed subspace M with {0} = M = H such that T M is a subset of M? Why it mattersA proof would guarantee that every such operator has a smaller closed space on which its action can be studied. A counterexample would show that even this basic form of reduction can fail. Status Open Status review date not recorded in this edition Sources Research on ProofAtlas Problem familyInvariant Subspace Problem for Hilbert Space Operators · 1 problem in this edition. Family membership alone does not assert an implication. - 108 · Invariant Subspace Problem for Hilbert Space Operators this problem Ranking uncertainty90% source-fit bootstrap rank range: 93–132. Source-fit opponent count: 35. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 108. Rank 109 Exact Kissing Numbers in Higher Dimensions In every dimension greater than four, what is the maximum number of equal unit spheres that can touch one central unit sphere without overlapping one another?Cryptography, coding, information & optimization Exact statementFor every integer n 4, determine the exact kissing number tau n: the maximum number of pairwise nonoverlapping unit balls in Euclidean R^n that can touch a central unit ball of the same radius. Why it mattersExact kissing numbers would determine the largest possible local contact arrangement in each dimension. A complete answer needs both an arrangement attaining the number and a proof that more contacts are impossible. Status Open · subcases solved Open with solved instances n=8 and n=24 within the target range; the March 2026 primary source still records a gap between 40 and 44 in dimension five. No full-sequence resolution was located. October 6, 2026: A four-dimensional packing-density/24-cell claim does not determine exact kissing numbers in every dimension n 4. The complete proof was not audited in this review. Status reviewed Sources - Musin, The kissing number in four dimensions https://arxiv.org/abs/math/0309430 Statement source - Status source for Exact Kissing Numbers in Higher Dimensions https://arxiv.org/abs/math/0309430 Status source - Status source for Exact Kissing Numbers in Higher Dimensions https://arxiv.org/html/2412.00937v3 Status source - Exact Kissing Numbers in Higher Dimensions: October 2026 scope review https://arxiv.org/abs/2609.25120 Status source · Retrieved Oct 6, 2026 Problem familyExact Kissing Numbers in Higher Dimensions · 1 problem in this edition. Family membership alone does not assert an implication. - 109 · Exact Kissing Numbers in Higher Dimensions this problem Ranking uncertainty90% source-fit bootstrap rank range: 94–128. Source-fit opponent count: 24. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 109. Rank 110 Unbounded Ranks of Elliptic Curves over Q For every nonnegative integer r, does some elliptic curve over the rational numbers have at least r independent rational points of infinite order?Number theory & arithmetic geometry Exact statementProve that for EVERY nonnegative integer r there is SOME elliptic curve E/Q whose actual Mordell-Weil group E Q has Z-rank at least r. The base field is exactly Q and rank means the free abelian rank in Mordell's theorem, not analytic rank, a Selmer bound or random-matrix corank. Thus the set of actual ranks of ALL elliptic curves over Q is unbounded. Counting one representative per Q-isomorphism class does not alter this existence question. No effective construction, specified family, uniform curve-producing algorithm or predetermined rate of growth is required. Why it mattersA proof would show that the number of independent generators needed for rational points has no universal ceiling across elliptic curves over the rationals. It would establish existence without requiring an algorithm that constructs the curves. Status Open Status review date not recorded in this edition Sources Research on ProofAtlas Problem familyUnbounded Ranks of Elliptic Curves over Q · 1 problem in this edition. Family membership alone does not assert an implication. - 110 · Unbounded Ranks of Elliptic Curves over Q this problem Ranking uncertainty90% source-fit bootstrap rank range: 98–130. Source-fit opponent count: 36. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 110. Rank 111 The Haldane Conjecture on Spectral Gaps of Antiferromagnetic Quantum Spin Chains In the one-dimensional isotropic antiferromagnetic Heisenberg spin chain, is there a positive energy gap above the ground state for every positive integer spin, but no gap for every half-integer spin?Mathematical physics Exact statementProve that the one-dimensional isotropic antiferromagnetic Heisenberg chain has a positive spectral gap for every integer spin S = 1 and is gapless for every half-integer spin. Why it mattersThis would establish a precise distinction in the low-energy behavior of the two spin classes: an energy threshold for excitations in integer-spin chains and excitations arbitrarily close to the ground-state energy in half-integer chains. Status Open · subcases solved Status review date not recorded in this edition Sources Problem familyThe Haldane Conjecture on Spectral Gaps of Antiferromagnetic Quantum Spin Chains · 1 problem in this edition. Family membership alone does not assert an implication. - 111 · The Haldane Conjecture on Spectral Gaps of Antiferromagnetic Quantum Spin Chains this problem Ranking uncertainty90% source-fit bootstrap rank range: 93–136. Source-fit opponent count: 28. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 111. Rank 112 Quantum Unique Ergodicity on Negatively Curved Manifolds On a compact negatively curved space, must every sequence of normalized Laplace eigenfunctions—standing-wave patterns—whose energies tend to infinity spread uniformly in both position and direction?Probability, ergodic theory & dynamics Exact statementLet M,g be a compact negatively curved Riemannian manifold. Prove that every complete high-energy sequence of L2-normalized Laplace eigenfunctions has semiclassical measures converging to Liouville measure on the unit cosphere bundle; consequently, the position densities converge weakly to normalized Riemannian volume. Why it mattersA proof would exclude exceptional high-energy sequences that retain concentration in phase space. It would also force their spatial probability densities to approach normalized volume throughout the manifold. Status Open Status review date not recorded in this edition Sources Problem familyQuantum Unique Ergodicity on Negatively Curved Manifolds · 1 problem in this edition. Family membership alone does not assert an implication. - 112 · Quantum Unique Ergodicity on Negatively Curved Manifolds this problem Ranking uncertainty90% source-fit bootstrap rank range: 95–139. Source-fit opponent count: 28. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 112. Rank 113 Iitaka conjecture C {n,m} on subadditivity of Kodaira dimension When one smooth projective variety is mapped algebraically onto another, is its Kodaira dimension—a measure of algebraic complexity—at least the sum of the base's Kodaira dimension and that of a general fibre?Geometry & topology Exact statementProve the Iitaka conjecture C {n,m}: for a surjective morphism f: X - Y between smooth projective varieties with general fibre F, kappa X = kappa Y + kappa F . Why it mattersKodaira dimension measures the growth of pluricanonical forms. The inequality would let information about the base and typical fibre give a lower bound for this measure on the total space. Status Open · subcases solved Status review date not recorded in this edition Sources Problem familyIitaka conjecture C {n,m} on subadditivity of Kodaira dimension · 1 problem in this edition. Family membership alone does not assert an implication. - 113 · Iitaka conjecture C {n,m} on subadditivity of Kodaira dimension this problem Ranking uncertainty90% source-fit bootstrap rank range: 93–141. Source-fit opponent count: 26. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 113. Rank 114 Prove the BKL locality conjecture in the general inhomogeneous context Near a singularity in generic gravitational collapse, do different spatial points evolve increasingly independently, following the oscillations of the spatially homogeneous vacuum Bianchi IX model, with the singularity forming a spacelike boundary?Mathematical physics Exact statementProve that the singularity in generic inhomogeneous gravitational collapse is spacelike, local, and oscillatory, and that the evolution at each spatial point approaches vacuum homogeneous Bianchi IX Mixmaster behavior. Why it mattersA proof would justify describing the approach to these singularities through local oscillatory dynamics despite spatial inhomogeneity. Failure of locality or of the proposed limiting behavior would require a different description of generic collapse. Status Open Status review date not recorded in this edition Sources Problem familyProve the BKL locality conjecture in the general inhomogeneous context · 1 problem in this edition. Family membership alone does not assert an implication. - 114 · Prove the BKL locality conjecture in the general inhomogeneous context this problem Ranking uncertainty90% source-fit bootstrap rank range: 87–157. Source-fit opponent count: 23. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 114. Rank 115 Erdos-Szemeredi Sum-Product Conjecture for Integers As finite sets of integers grow, must either their distinct pairwise sums or their distinct pairwise products number almost quadratically many—at least a constant times the set size to the power 2−ε, for every ε 0, with the constant depending only on ε?Combinatorics & discrete geometry Exact statementProve that for every epsilon greater than zero there is a constant c epsilon greater than zero such that every finite set A of integers satisfies max |A+A|, |A A| at least c epsilon times |A|^ 2-epsilon . Why it mattersA proof would show that a large integer set cannot have both highly compressed addition and highly compressed multiplication. At least one operation would generate almost quadratically many distinct outcomes. Status Open October 6, 2026: Counterexamples using real algebraic integers do not refute the stated target for subsets of ordinary integers Z. The earlier arXiv:2605.28781 also concerns real numbers. The complete proof was not audited in this review. Status reviewed Sources Problem familyErdos-Szemeredi Sum-Product Conjecture for Integers · 1 problem in this edition. Family membership alone does not assert an implication. - 115 · Erdos-Szemeredi Sum-Product Conjecture for Integers this problem Ranking uncertainty90% source-fit bootstrap rank range: 103–136. Source-fit opponent count: 27. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 115. Rank 116 Regular Inverse Galois Problem over Q Can every finite group occur as the Galois group of a finite Galois extension of Q T whose only elements algebraic over Q are rational numbers? This asks for a realization over a rational function field without introducing new algebraic constants.Algebra, representation & category theory Inverse Galois over Q · 2 problems family-context-problem-regular-inverse-galois-problem-over-q Exact statementFor every finite group G, does there exist a finite Galois extension F/Q T with Gal F/Q T isomorphic to G and Q algebraically closed in F? Equivalently, require a geometrically integral Galois cover of the projective line over Q with group G. A positive resolution proves this for all finite groups; a negative resolution proves nonexistence for at least one finite group. Why it mattersSuch realizations would supply geometric covers for every finite group. Hilbert irreducibility would then give every finite group as a Galois group over Q, settling ordinary inverse Galois affirmatively. Status Open · subcases solved Status reviewed Sources - Demeio and Gvirtz-Chen, Galois Realisations of PSL2 Fp^2 https://arxiv.org/html/2512.23615v1 Statement source Problem familyInverse Galois over Q · 2 problems in this edition. Family membership alone does not assert an implication. - 60 · Inverse Galois problem over the rational numbers problem-inverse-galois-problem-over-the-rational-numbers - 116 · Regular Inverse Galois Problem over Q this problem Ranking uncertainty90% source-fit bootstrap rank range: 101–139. Source-fit opponent count: 30. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 116. Rank 117 The Borel Conjecture on Topological Rigidity of Aspherical Manifolds For closed manifolds whose universal covers contract to a point, can every homotopy equivalence—an equivalence up to continuous deformation—itself be deformed into a homeomorphism, a continuous bijection with continuous inverse?Geometry & topology Exact statementProve that every homotopy equivalence between closed aspherical manifolds is homotopic to a homeomorphism. Why it mattersA proof would make homotopy information sufficient to recover the topology of these manifolds, while also controlling the particular map that relates them. A counterexample would expose a distinction invisible to homotopy equivalence. Status Open · subcases solved Status review date not recorded in this edition Sources Research on ProofAtlas Problem familyThe Borel Conjecture on Topological Rigidity of Aspherical Manifolds · 1 problem in this edition. Family membership alone does not assert an implication. - 117 · The Borel Conjecture on Topological Rigidity of Aspherical Manifolds this problem Ranking uncertainty90% source-fit bootstrap rank range: 94–157. Source-fit opponent count: 23. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 117. Rank 118 Grothendieck–Katz p-Curvature Conjecture An integrable algebraic connection describes a compatible system of differential equations. If its p-curvature vanishes after reduction at almost every prime, must it have a complete set of algebraic solutions?Number theory & arithmetic geometry Exact statementLet X be a smooth connected complex variety and V,∇ an algebraic vector bundle with integrable connection, spread out over a finitely generated Z-algebra. Prove that if the reductions have zero p-curvature for almost all primes, then V,∇ has a full set of algebraic solutions, equivalently becomes trivial on a finite étale cover and has finite monodromy. Why it mattersA proof would make arithmetic information from reductions modulo primes sufficient to guarantee algebraic solvability over the complex numbers. Equivalently, the connection would become trivial on a finite étale cover and have finite monodromy. Status Open · subcases solved Status review date not recorded in this edition Sources Research on ProofAtlas Problem familyGrothendieck–Katz p-Curvature Conjecture · 1 problem in this edition. Family membership alone does not assert an implication. - 118 · Grothendieck–Katz p-Curvature Conjecture this problem Ranking uncertainty90% source-fit bootstrap rank range: 100–145. Source-fit opponent count: 20. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 118. Rank 119 Serre's congruence subgroup conjecture For arithmetic groups arising from simply connected, absolutely almost simple algebraic groups, is the congruence kernel—the gap between finite-quotient and congruence information—finite in total rank at least two when all selected nonarchimedean local ranks are positive, but infinite in rank one?Algebra, representation & category theory Exact statementLet K be a global field a number field or the function field of a curve over a finite field , G an absolutely almost simple simply connected algebraic K-group, and S a finite nonempty set of places of K containing every archimedean place when K is a number field. Put r S G = sum over v in S of rank {K v} G , and let C^S G be the kernel of the natural surjection from the profinite completion of G O S S-arithmetic topology onto its S-congruence completion. Prove Serre's congruence subgroup conjecture: a if r S G = 2 and rank {K v} G 0 for every nonarchimedean v in S, then C^S G is finite; b if r S G = 1, then C^S G is infinite. The case r S G = 0 finite G O S , trivial kernel is excluded. Finiteness of C^S G is the assertion; triviality is a stronger statement not required. Which specific groups remain unsettled is status information, not part of the target. Why it mattersThe kernel measures the discrepancy between two ways of recovering an arithmetic group from finite information. A proof would classify that discrepancy as finite or infinite under these rank hypotheses, without requiring finite kernels to be trivial. Status Open · subcases solved Status review date not recorded in this edition Sources Problem familySerre's congruence subgroup conjecture · 1 problem in this edition. Family membership alone does not assert an implication. - 119 · Serre's congruence subgroup conjecture this problem Ranking uncertainty90% source-fit bootstrap rank range: 105–141. Source-fit opponent count: 34. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 119. Rank 120 Cohen-Lenstra class-group distributions for quadratic number fields Class groups measure how unique factorization can fail in number fields. Among quadratic fields ordered by discriminant size, do their odd-prime parts follow the Cohen–Lenstra frequency law, favoring groups with fewer symmetries and giving smaller groups extra weight in the real case?Number theory & arithmetic geometry Exact statementFor u in {0,1} let K u X be the set of isomorphism classes of quadratic number fields K with |disc K | < X that are imaginary u = 0 or real u = 1 . For each fixed odd prime p and each finite abelian p-group A, prove that the limit as X tends to infinity of {K in K u X : Cl K p^infinity is isomorphic to A} / K u X exists and equals 1/Z u p |A|^{-u} / |Aut A |, where Z u p is the sum over isomorphism classes of finite abelian p-groups B of |B|^{-u}/|Aut B | so Z 0 p = prod {i =1} 1-p^{-i} ^{-1} and Z 1 p = prod {i =2} 1-p^{-i} ^{-1} . Cl K is the ordinary ideal class group and Cl K p^infinity its Sylow p-subgroup. No assertion is made for p = 2, for narrow class groups, for a joint law over several primes, or for any non-quadratic family. Why it mattersThe formula would specify the limiting frequency of every finite abelian p-group in each quadratic family. It would turn a weighting rule into exact statistical predictions for ordinary class groups, separately for real and imaginary fields. Status Open Status review date not recorded in this edition Sources Problem familyCohen-Lenstra class-group distributions for quadratic number fields · 1 problem in this edition. Family membership alone does not assert an implication. - 120 · Cohen-Lenstra class-group distributions for quadratic number fields this problem Ranking uncertainty90% source-fit bootstrap rank range: 105–143. Source-fit opponent count: 27. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 120. Rank 121 QMA versus QCMA Problem Can every decision problem checked efficiently by a quantum computer using a polynomial-size quantum certificate also be checked efficiently using only a polynomial-size classical certificate?Quantum information & computation Quantum verification classes · 2 problems family-context-problem-qma-versus-qcma-problem Exact statementDetermine whether QMA=QCMA: does every language decidable by a polynomial-time quantum verifier using a polynomial-size quantum witness also admit a polynomial-time quantum verification protocol using only a polynomial-size classical witness? Why it mattersThe answer would determine whether quantum states give witnesses more verification power than classical strings when the verifier is already quantum. Equality would permit classical witnesses for all languages in the quantum-witness class. Status Open · subcases solved Status review date not recorded in this edition Sources Problem familyQuantum verification classes · 2 problems in this edition. Family membership alone does not assert an implication. - 121 · QMA versus QCMA Problem this problem - 266 · Equivalence or Strict Containment of QMA and QMA 2 problem-equivalence-or-strict-containment-of-qma-and-qma-2 Ranking uncertainty90% source-fit bootstrap rank range: 92–168. Source-fit opponent count: 25. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 121. Rank 122 Volume conjecture for knots A hyperbolic knot has both polynomial invariants and a geometric volume for the space around it. Does that volume equal 2π times the exponential growth rate of the absolute values of its normalized colored Jones polynomials, evaluated at e^ 2πi/N as the color N grows?Geometry & topology Exact statementProve or disprove that for every hyperbolic knot K in S^3, the limit as N tends to infinity of 2 pi/N log|J {K,N} exp 2 pi i/N | equals the hyperbolic volume of S^3 minus K, using the standard normalized N-colored Jones polynomial. Why it mattersA proof would recover a geometric volume from a sequence of polynomial knot invariants. A counterexample would locate a failure of that proposed link, either through a different limit or a limit that does not exist. Status Open · subcases solved Status review date not recorded in this edition Sources Research on ProofAtlas Problem familyVolume conjecture for knots · 1 problem in this edition. Family membership alone does not assert an implication. - 122 · Volume conjecture for knots this problem Ranking uncertainty90% source-fit bootstrap rank range: 103–151. Source-fit opponent count: 23. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 122. Rank 123 Aspherical Manifolds Admit No Metric of Positive Scalar Curvature Can a closed manifold whose universal cover contracts to a point ever carry a Riemannian metric with scalar curvature positive everywhere?Geometry & topology Exact statementProve or disprove that no closed aspherical manifold admits a Riemannian metric of positive scalar curvature. Why it mattersA proof would make asphericity a universal topological obstruction to positive scalar curvature. A counterexample would exhibit a manifold where this topology and everywhere positive scalar curvature can coexist. Status Open · subcases solved Status review date not recorded in this edition Sources Problem familyAspherical Manifolds Admit No Metric of Positive Scalar Curvature · 1 problem in this edition. Family membership alone does not assert an implication. - 123 · Aspherical Manifolds Admit No Metric of Positive Scalar Curvature this problem Ranking uncertainty90% source-fit bootstrap rank range: 102–158. Source-fit opponent count: 27. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 123. Rank 124 Erdos-Hajnal Conjecture If a graph avoids one fixed induced pattern, including its absent edges, must it contain vertices that are either all mutually connected or all mutually disconnected in a set of size at least a fixed positive power of the total number of vertices?Combinatorics & discrete geometry Exact statementProve that for every fixed finite graph H there is a constant c H 0 such that every n-vertex graph with no induced subgraph isomorphic to H contains a clique or an independent set of size at least n^{c H}. Why it mattersA proof would force a large completely ordered pattern of adjacency whenever one induced pattern is excluded. The guaranteed size would be a power of the total vertex count, with the exponent depending only on the forbidden graph. Status Open · subcases solved Status review date not recorded in this edition Sources Research on ProofAtlas Problem familyErdos-Hajnal Conjecture · 1 problem in this edition. Family membership alone does not assert an implication. - 124 · Erdos-Hajnal Conjecture this problem Ranking uncertainty90% source-fit bootstrap rank range: 106–153. Source-fit opponent count: 28. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 124. Rank 125 Density Hypothesis for Zeroes of the Riemann Zeta Function How sparse are zeta zeros to the right of the critical line? For 1/2≤σ≤1, must their count with real part at least σ and imaginary part of magnitude at most T grow no faster than a constant times T^ 2 1−σ +ε , for every ε 0?Number theory & arithmetic geometry Zeta zeros and large Dirichlet polynomial values · 2 problems family-context-problem-density-hypothesis-for-zeroes-of-the-riemann-zeta-function Exact statementProve that for every sigma with 1/2 <= sigma <= 1 and every epsilon 0, the number N sigma,T of nontrivial zeros rho = beta + i gamma of the Riemann zeta function with beta = sigma and |gamma| <= T satisfies N sigma,T <= C {sigma,epsilon} T^{2 1-sigma +epsilon} for all sufficiently large T. Why it mattersThe estimate would quantify how sparse zeros can be toward the right of the critical strip. Such control feeds estimates for the distribution of prime numbers, including primes in short intervals. Status Open Status review date not recorded in this edition Sources Problem familyZeta zeros and large Dirichlet polynomial values · 2 problems in this edition. Family membership alone does not assert an implication. - 125 · Density Hypothesis for Zeroes of the Riemann Zeta Function this problem - 308 · Montgomery's Large Value Conjecture for Dirichlet Polynomials problem-montgomery-s-large-value-conjecture-for-dirichlet-polynomials Ranking uncertainty90% source-fit bootstrap rank range: 95–184. Source-fit opponent count: 21. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 125. Rank 126 No-Slip Vanishing-Viscosity Limit in the Energy Norm As viscosity tends to zero in a smooth bounded two- or three-dimensional container with no-slip walls, must every finite-energy Leray–Hopf flow approach the ideal Euler flow with mean-square velocity error tending uniformly to zero on every time interval where Euler stays smooth, when both start from the same smooth boundary-vanishing incompressible velocity?Analysis & PDE Exact statementFor d in {2,3}, every smooth bounded domain Omega in R^d and every fixed smooth divergence-free initial velocity u0 vanishing on the boundary, prove or disprove that every no-slip Leray–Hopf Navier–Stokes family u^nu with viscosity nu and initial value u0 converges to the Euler solution u in L^infinity 0,T ;L^2 Omega as nu tends to zero, for each T on which u is smooth and satisfies u.n=0. Why it mattersA proof would justify this energy-norm limit on each interval where the impermeable-boundary Euler solution remains smooth. A counterexample would show that vanishing viscosity can retain a nonvanishing L² difference under these conditions. Status Open Status review date not recorded in this edition Sources Problem familyNo-Slip Vanishing-Viscosity Limit in the Energy Norm · 1 problem in this edition. Family membership alone does not assert an implication. - 126 · No-Slip Vanishing-Viscosity Limit in the Energy Norm this problem Ranking uncertainty90% source-fit bootstrap rank range: 107–154. Source-fit opponent count: 21. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 126. Rank 127 Local Connectivity of the Mandelbrot Set MLC Conjecture At every point of the Mandelbrot set—the parameters c for which iterating z ↦ z² + c from zero stays bounded—are there arbitrarily small connected neighborhoods within the set?Probability, ergodic theory & dynamics Mandelbrot set boundary · 2 problems family-context-problem-local-connectivity-of-the-mandelbrot-set-mlc-conjecture Exact statementDetermine whether the Mandelbrot set M = {c in C : the orbit of 0 under z - z^2 + c is bounded} is locally connected at every point. Why it mattersA proof would control the set's local topology even at its most intricate parameters. A counterexample would identify a point where connected neighborhoods cannot be made arbitrarily small. Status Open · subcases solved Status review date not recorded in this edition Sources Problem familyMandelbrot set boundary · 2 problems in this edition. Family membership alone does not assert an implication. - 127 · Local Connectivity of the Mandelbrot Set MLC Conjecture this problem - 410 · Lebesgue measure of the boundary of the Mandelbrot set problem-top500-omission-w050054-lebesgue-measure-of-the-boundary-of-the-mandelbrot-set Ranking uncertainty90% source-fit bootstrap rank range: 106–157. Source-fit opponent count: 25. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 127. Rank 128 Polynomial Bound for Limit Cycles of Planar Polynomial Vector Fields Can the number of isolated periodic motions in any polynomial differential system in the plane be bounded by a power of its degree, with one universal exponent?Probability, ergodic theory & dynamics Exact statementDetermine whether the number K of limit cycles of every planar polynomial differential system of degree at most d satisfies K ≤ d^q for some universal exponent q. Why it mattersSuch a bound would control the number of isolated repeating motions directly from the degree of the equations. A negative answer would show that no fixed polynomial in the degree captures their possible number. Status Open Status review date not recorded in this edition Sources - Steve Smale, Mathematical Problems for the Next Century, 1998. https://www.fim.uni-passau.de/fileadmin/dokumente/fakultaeten/fim/lehrstuhl/muller/SmaleProblems1998.pdf Statement source Problem familyPolynomial Bound for Limit Cycles of Planar Polynomial Vector Fields · 1 problem in this edition. Family membership alone does not assert an implication. - 128 · Polynomial Bound for Limit Cycles of Planar Polynomial Vector Fields this problem Ranking uncertainty90% source-fit bootstrap rank range: 91–175. Source-fit opponent count: 22. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 128. Rank 129 Tarski's Exponential Function Problem Can an algorithm always decide whether any first-order statement about real numbers, using arithmetic, inequalities and the exponential function, is true—even when it quantifies over all real numbers?Logic, foundations & set theory Exact statementDetermine whether the first-order theory of the real exponential field R; 0, 1, +, , exp, <, = is decidable. Why it mattersA positive answer would provide a universal decision procedure for this precisely specified language, including nested quantifiers over real numbers. A negative answer would prove that no algorithm can decide all such sentences. Status Open Status review date not recorded in this edition Sources Research on ProofAtlas Problem familyTarski's Exponential Function Problem · 1 problem in this edition. Family membership alone does not assert an implication. - 129 · Tarski's Exponential Function Problem this problem Ranking uncertainty90% source-fit bootstrap rank range: 107–157. Source-fit opponent count: 40. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 129. Rank 130 Unconditional Single-Prover Classical Verification of BQP Computations Can an ordinary computer efficiently check any efficient quantum computation by exchanging classical messages with one quantum prover, reliably accepting correct answers and rejecting false ones even if a cheating prover has unlimited computing power?Quantum information & computation Exact statementDetermine whether every language L in BQP admits a polynomial-round interactive proof with only classical messages between a probabilistic classical polynomial-time verifier and a single honest uniform quantum polynomial-time prover, with completeness at least 2/3 and soundness at most 1/3 against every computationally unbounded cheating prover. Why it mattersAn affirmative answer would give purely classical verification of every computation in BQP while requiring only an efficient honest quantum prover. The guarantee would resist computationally unbounded cheating without relying on cryptographic hardness assumptions. Status Open · subcases solved Status review date not recorded in this edition Sources Problem familyUnconditional Single-Prover Classical Verification of BQP Computations · 1 problem in this edition. Family membership alone does not assert an implication. - 130 · Unconditional Single-Prover Classical Verification of BQP Computations this problem Ranking uncertainty90% source-fit bootstrap rank range: 101–177. Source-fit opponent count: 20. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 130. Rank 131 Existence of a Length-Optimal Propositional Proof System Is there a Cook–Reckhow propositional proof system whose shortest proofs are at most polynomially longer than those in any other such system, with the polynomial allowed to depend on the other system? An efficient way to translate proofs is not required.Theoretical computer science Optimal propositional proof systems · 2 problems family-context-problem-existence-of-length-optimal-propositional-proof-system Exact statementDoes there exist a polynomial-time computable function P from finite binary strings onto the encodings of all propositional tautologies such that, for every polynomial-time computable function Q onto the same tautologies, there is a polynomial q Q for which every Q-proof pi has some P-proof sigma with P sigma =Q pi and |sigma| <= q Q |pi| ? The polynomial may depend on Q. No polynomial-time algorithm for finding sigma from pi is required. This is the ordinary unrelativized Cook–Reckhow setting without advice or randomness; resolve existence or nonexistence of such P. Also known asOptimal propositional proof system problem; Length-optimal Cook–Reckhow proof system; Existence of an optimal proof system for TAUT Why it mattersThis asks whether one proof system can match every other system in proof length. It separates the existence of short proofs from the algorithmic task of finding or translating them. Status Open Status reviewed Sources - Recursive Jump Operators and Optimal Proof Systems https://drops.dagstuhl.de/storage/00lipics/lipics-vol374-icalp2026/html/LIPIcs.ICALP.2026.88/LIPIcs.ICALP.2026.88.html Statement source Problem familyOptimal propositional proof systems · 2 problems in this edition. Family membership alone does not assert an implication. - 131 · Existence of a Length-Optimal Propositional Proof System this problem - 361 · Existence of a p-Optimal Propositional Proof System problem-existence-of-a-p-optimal-propositional-proof-system Ranking uncertainty90% source-fit bootstrap rank range: 109–162. Source-fit opponent count: 24. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 131. Rank 132 Polynomial pivot rule for the simplex method Can a rule for choosing the simplex method's next step solve every bounded nondegenerate rational linear program using polynomially many pivots in its numbers of variables and constraints, from any feasible starting basis? A randomized rule may meet this bound in expectation.Cryptography, coding, information & optimization Exact statementConstruct a deterministic or randomized simplex pivot rule and a polynomial p such that, for every bounded nondegenerate rational linear program with m equality constraints and n variables and every feasible starting basis, the rule reaches an optimal basis in at most p m,n pivots, in expectation when the rule is randomized; or prove that no such rule exists. Why it mattersA construction would bound the number of steps along feasible edges using only the numbers of constraints and variables. It would control pivot count without, by itself, bounding the computational work needed to choose each pivot. Status Open Status review date not recorded in this edition Sources - Math Conjectures, OPT-002, catalog snapshot 31 July 2026. https://mathconjectures.com/conjectures/OPT-002 Statement source Problem familyPolynomial pivot rule for the simplex method · 1 problem in this edition. Family membership alone does not assert an implication. - 132 · Polynomial pivot rule for the simplex method this problem Ranking uncertainty90% source-fit bootstrap rank range: 108–173. Source-fit opponent count: 26. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 132. Rank 133 The Generalized Sato-Tate Conjecture for Higher-Dimensional Abelian Varieties As primes vary, do the normalized Frobenius data of every abelian variety over a number field, encoding its behavior modulo those primes, follow the probability distribution predicted by its Sato–Tate symmetry group?Number theory & arithmetic geometry Exact statementProve that the normalized Frobenius conjugacy classes of an abelian variety A of dimension g over a number field K are equidistributed in the Sato-Tate group ST A . Why it mattersA proof would give a precise statistical law for these arithmetic data across primes. The Sato–Tate group would determine the limiting distribution rather than merely describe possible individual Frobenius classes. Status Open Status review date not recorded in this edition Sources Research on ProofAtlas Problem familyThe Generalized Sato-Tate Conjecture for Higher-Dimensional Abelian Varieties · 1 problem in this edition. Family membership alone does not assert an implication. - 133 · The Generalized Sato-Tate Conjecture for Higher-Dimensional Abelian Varieties this problem Ranking uncertainty90% source-fit bootstrap rank range: 106–181. Source-fit opponent count: 20. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 133. Rank 134 Erdős problem 3: divergent harmonic series and long arithmetic progressions If the reciprocals of the positive integers in a set have an infinite sum, must the set contain arbitrarily long sequences of equally spaced integers?Number theory & arithmetic geometry Exact statementDetermine whether every subset A of the natural numbers whose reciprocal sum diverges must contain arbitrarily long arithmetic progressions. Why it mattersA proof would turn the divergence of one numerical sum into a guarantee of arbitrarily long additive patterns. A counterexample would separate this measure of abundance from the existence of long arithmetic progressions. Status Open · subcases solved Status review date not recorded in this edition Sources Research on ProofAtlas Problem familyErdős problem 3: divergent harmonic series and long arithmetic progressions · 1 problem in this edition. Family membership alone does not assert an implication. - 134 · Erdős problem 3: divergent harmonic series and long arithmetic progressions this problem Ranking uncertainty90% source-fit bootstrap rank range: 118–166. Source-fit opponent count: 53. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 134. Rank 135 NP-Completeness of the Minimum Circuit Size Problem MCSP Given a function's full table of binary inputs and outputs, is deciding whether it has a circuit of at most a chosen size as hard as every problem in NP, under deterministic polynomial-time many-one reductions?Theoretical computer science Exact statementProve or disprove that MCSP is NP-hard under deterministic polynomial-time many-one reductions, where an input is the truth table of a Boolean function and a size bound and asks whether a circuit of at most that size computes the function. Why it mattersA hardness proof would make deciding whether a compact circuit description exists a destination for reductions from every NP problem. A negative answer would distinguish this decision problem from that standard form of universal NP hardness. Status Open Status review date not recorded in this edition Sources Problem familyNP-Completeness of the Minimum Circuit Size Problem MCSP · 1 problem in this edition. Family membership alone does not assert an implication. - 135 · NP-Completeness of the Minimum Circuit Size Problem MCSP this problem Ranking uncertainty90% source-fit bootstrap rank range: 122–157. Source-fit opponent count: 43. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 135. Rank 136 Arithmetic Quantum Unique Ergodicity On arithmetic locally symmetric spaces, must high-energy wave patterns that are also Hecke eigenfunctions spread uniformly in the limit, leaving normalized volume as their only possible limiting distribution?Probability, ergodic theory & dynamics Exact statementProve the arithmetic quantum unique ergodicity conjecture: for arithmetic locally symmetric spaces, the normalized Riemannian volume is the only arithmetic quantum limit of Hecke eigenfunctions. Why it mattersA proof would force these arithmetically constrained eigenfunctions to spread uniformly in the limit. It would exclude alternative limiting distributions that retain concentration in particular regions of the space. Status Open · subcases solved Status review date not recorded in this edition Sources Problem familyArithmetic Quantum Unique Ergodicity · 1 problem in this edition. Family membership alone does not assert an implication. - 136 · Arithmetic Quantum Unique Ergodicity this problem Ranking uncertainty90% source-fit bootstrap rank range: 111–174. Source-fit opponent count: 30. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 136. Rank 137 Borel normal number conjecture for irrational algebraic numbers Must every real irrational root of an integer polynomial have equally frequent digit blocks of each fixed length in every integer base of at least two?Number theory & arithmetic geometry Exact statementProve or disprove that every real irrational algebraic number is normal to every integer base b = 2. Why it mattersA proof would impose precise long-run frequencies on all digit patterns of algebraic irrationals. It would connect numbers defined by polynomial equations to the same block frequencies as uniformly distributed random digits. Status Open Status review date not recorded in this edition Sources - Michel Waldschmidt, Words and Transcendence, 2005. https://webusers.imj-prg.fr/~michel.waldschmidt/articles/pdf/WordsTranscendence.pdf Statement source Problem familyBorel normal number conjecture for irrational algebraic numbers · 1 problem in this edition. Family membership alone does not assert an implication. - 137 · Borel normal number conjecture for irrational algebraic numbers this problem Ranking uncertainty90% source-fit bootstrap rank range: 107–182. Source-fit opponent count: 20. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 137. Rank 138 Hyperlinearity of Every Group Can every finite piece of any group's multiplication table be modeled arbitrarily accurately by unitary matrices, while nonidentity elements stay separated from the identity in the normalized Hilbert–Schmidt distance?Algebra, representation & category theory Group approximation and determinants · 2 problems family-context-problem-top500-omission-w045049-hyperlinearity-of-every-group Exact statementDetermine whether every group G is hyperlinear. Precisely, for each finite F contained in G and each epsilon 0, must there be d =1 and phi:F- U d such that ||phi g phi h -phi gh || 2,d <epsilon whenever g,h,gh belong to F, and ||phi g -I|| 2,d sqrt 2 -epsilon for every nonidentity g in F? Here ||X|| 2,d = Tr X X /d ^ 1/2 is the normalized Hilbert-Schmidt norm. A negative resolution is a group failing this condition. Why it mattersA proof would give finite-dimensional unitary approximations for every group in the precise stated sense. A counterexample would identify a finite pattern and tolerance for which the required approximation and separation cannot coexist. Status Open Status review date not recorded in this edition Sources Problem familyGroup approximation and determinants · 2 problems in this edition. Family membership alone does not assert an implication. - 138 · Hyperlinearity of Every Group this problem - 420 · Lück's Determinant Conjecture for Discrete Groups problem-top500-omission-w055056-luck-determinant Ranking uncertainty90% source-fit bootstrap rank range: 112–175. Source-fit opponent count: 21. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 138. Rank 139 Planted Clique Conjecture Hide a complete subgraph on about n^ 1/2−α randomly chosen vertices in an otherwise random n-vertex graph whose edges appear independently with probability 1/2. For every fixed 0<α<1/2, is recovering all hidden vertices with probability tending to one impossible in randomized polynomial time?Theoretical computer science Exact statementFor every fixed real alpha with 0<alpha<1/2, set k n=floor n^ 1/2-alpha . Sample a uniform k n-subset S of n labeled vertices, put every edge inside S, and independently include each other edge with probability1/2. There is no randomized algorithm running in time polynomial in n that, given this graph, outputs exactly S with probability tending to1 as n tends to infinity, over S, remaining edges and its internal coins. Algorithms may depend on the fixed alpha but receive no advice depending on n. Why it mattersA proof would establish a computational barrier to recovering the actual planted vertex set in this random-graph model. An algorithm refuting it would need exact recovery for some fixed exponent, rather than merely detecting that a clique was planted. Status Open · subcases solved Status review date not recorded in this edition Sources - Planted Clique Conjectures Are Equivalent, ECCC TR24-058,29March2024 https://eccc.weizmann.ac.il/report/2024/058/download/ Statement source Problem familyPlanted Clique Conjecture · 1 problem in this edition. Family membership alone does not assert an implication. - 139 · Planted Clique Conjecture this problem Ranking uncertainty90% source-fit bootstrap rank range: 114–173. Source-fit opponent count: 22. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 139. Rank 140 Global Gan–Gross–Prasad Conjecture Fixed-Representation Form For a fixed tempered cuspidal automorphic representation in the Gan–Gross–Prasad setting of classical groups and their covers, with multiplicity one, is its global period nonzero exactly when every local compatibility test succeeds and its associated L-function has a nonzero central value?Number theory & arithmetic geometry Exact statementFor every number field F, let E be F, a quadratic extension, or F×F, and let W⊂V be a nondegenerate relevant pair with form sign epsilon, split W-perp, and epsilon -1 ^dim W-perp =-1; a two-dimensional orthogonal V or W is required to be nonsplit. Use exactly the groups, covers, Bessel/Fourier–Jacobi representation nu and auxiliary data of Gan–Gross–Prasad Section 23, and R of Section 24, including the skew-hermitian character mu and its matching inverse twist in R. For every irreducible tempered representation pi occurring with multiplicity one in the cuspidal automorphic spectrum, Conjecture 24.1 asserts: the global period F nu is nonzero on pi if and only if Hom H F v pi v,nu v is nonzero for every place v and L pi,R,1/2 is nonzero. Quantification is over all data permitted there, including relevant orthogonal, hermitian, symplectic/metaplectic and skew-hermitian cases. This is the fixed-representation equivalence, not the refined period identity or packet-existence formulation. Why it mattersThe equivalence would decide global period nonvanishing from local representation data and one central L-value. It would connect these arithmetic and representation-theoretic tests for each fixed representation, without asserting a refined formula for the period's size. Status Open · subcases solved Status review date not recorded in this edition Sources Problem familyGlobal Gan–Gross–Prasad Conjecture Fixed-Representation Form · 1 problem in this edition. Family membership alone does not assert an implication. - 140 · Global Gan–Gross–Prasad Conjecture Fixed-Representation Form this problem Ranking uncertainty90% source-fit bootstrap rank range: 124–166. Source-fit opponent count: 43. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 140. Rank 141 Torsion Conjecture for Abelian Varieties of Dimension at Least Two After fixing a number field and a dimension of at least two, is there one common bound on the number of finite-order points defined over that field on any abelian variety of that dimension?Number theory & arithmetic geometry Exact statementFor every integer g =2 and every number field K, there exists a finite constant B g,K such that every abelian variety A defined over K of dimension g satisfies A K tors <= B g,K . Here A K tors is the subgroup of K-rational torsion points. The bound is uniform over all such A, without dependence on A, its height, endomorphisms, polarization degree or reduction type; it may depend on g and K. No effective formula, explicit numerical value or classification of torsion groups is required. Why it mattersA proof would prevent rational torsion from growing arbitrarily large while dimension and field remain fixed. The bound could exist without an explicit formula or a classification of the possible torsion groups. Status Open · subcases solved Status review date not recorded in this edition Sources Problem familyTorsion Conjecture for Abelian Varieties of Dimension at Least Two · 1 problem in this edition. Family membership alone does not assert an implication. - 141 · Torsion Conjecture for Abelian Varieties of Dimension at Least Two this problem Ranking uncertainty90% source-fit bootstrap rank range: 119–177. Source-fit opponent count: 22. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 141. Rank 142 Cannon's conjecture If a hyperbolic group's boundary at infinity is a two-dimensional sphere, must the group act by distance-preserving symmetries of hyperbolic three-space, properly discontinuously and with compact quotient?Geometry & topology Exact statementProve or disprove that every Gromov-hyperbolic group whose boundary at infinity is homeomorphic to the two-sphere acts properly discontinuously, cocompactly, and isometrically on hyperbolic three-space, equivalently is virtually a cocompact Kleinian group. Why it mattersA proof would let the topology of a group's boundary determine the existence of a three-dimensional hyperbolic geometric realization. A counterexample would show that a spherical boundary alone does not force that realization. Status Open · subcases solved Status review date not recorded in this edition Sources - Expanding Thurston Maps https://www.ams.org/bookstore/pspdf/surv-225-prev.pdf Statement source Research on ProofAtlas Problem familyCannon's conjecture · 1 problem in this edition. Family membership alone does not assert an implication. - 142 · Cannon's conjecture this problem Ranking uncertainty90% source-fit bootstrap rank range: 118–169. Source-fit opponent count: 22. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 142. Rank 143 Indistinguishability Obfuscation from Constant-Degree Homomorphic Assumptions Can we efficiently disguise programs so that programs doing the same computation are computationally indistinguishable, using only standard polynomial-time hardness assumptions and without a subexponential security loss?Cryptography, coding, information & optimization Exact statementConstruct indistinguishability obfuscation based purely on standard polynomial-time assumptions such as standard LWE, LPN, and degree-2 multilinear pairings without sub-exponential loss. Why it mattersSuch a construction would connect the obfuscation guarantee to the specified assumptions without the stated security loss. Its security would still be conditional on those assumptions and would concern indistinguishability of equivalent programs. Status Open Status review date not recorded in this edition Sources - A. Jain, H. Lin, A. Sahai, STOC 2021: 60-73, 2021. https://doi.org/10.1145/3406325.3451093 Statement source Problem familyIndistinguishability Obfuscation from Constant-Degree Homomorphic Assumptions · 1 problem in this edition. Family membership alone does not assert an implication. - 143 · Indistinguishability Obfuscation from Constant-Degree Homomorphic Assumptions this problem Ranking uncertainty90% source-fit bootstrap rank range: 113–203. Source-fit opponent count: 21. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 143. Rank 144 Superlinear Circuit Lower Bounds for Explicit Boolean Functions Can a family of yes-or-no functions be computable in polynomial time yet require circuits whose number of two-input gates grows faster than any constant times the input length, even with unrestricted depth and fan-out?Theoretical computer science Exact statementThere exists a Boolean function family f n:{0,1}^n- {0,1}, n =1, evaluated by a single deterministic polynomial-time algorithm on 1^n,x , whose minimum gate count C B2 f n over single-output fan-in-two circuits using arbitrary binary Boolean gates satisfies C B2 f n =omega n : for every c 0 there is N such that C B2 f n c n for every n =N. The circuits have unrestricted depth and fan-out; no fixed linear constant counts as superlinear. Why it mattersA proof would exhibit an efficiently specified computation whose circuit size grows faster than every linear bound. The eventual growth requirement excludes merely improving a fixed constant in front of n or obtaining lower bounds at isolated input lengths. Status Open Status review date not recorded in this edition Sources Problem familySuperlinear Circuit Lower Bounds for Explicit Boolean Functions · 1 problem in this edition. Family membership alone does not assert an implication. - 144 · Superlinear Circuit Lower Bounds for Explicit Boolean Functions this problem Ranking uncertainty90% source-fit bootstrap rank range: 117–182. Source-fit opponent count: 32. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 144. Rank 145 Residual Finiteness of Hyperbolic Groups For every word-hyperbolic group and each element other than its identity, is there a homomorphism into some finite group that keeps that element distinct from the identity?Algebra, representation & category theory Exact statementFor every word-hyperbolic group G and every nonidentity element g in G, prove that there is a finite group F and a homomorphism φ:G→F with φ g ≠1. Why it mattersA proof would ensure that finite group images can detect every nonidentity element, although different elements may require different images. A counterexample would exhibit an element invisible in every finite quotient. Status Open · subcases solved Status review date not recorded in this edition Sources Problem familyResidual Finiteness of Hyperbolic Groups · 1 problem in this edition. Family membership alone does not assert an implication. - 145 · Residual Finiteness of Hyperbolic Groups this problem Ranking uncertainty90% source-fit bootstrap rank range: 121–182. Source-fit opponent count: 26. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 145. Rank 146 Uniform Subconvexity for Standard Automorphic L-Functions on GL n An L-function’s analytic conductor measures its arithmetic and analytic complexity. For each fixed rank n and number field, are the magnitudes of the central values at s = 1/2 of all unitary cuspidal standard L-functions on GL n bounded by a constant times the conductor raised to an exponent strictly below one-quarter, uniformly as all parameters vary together?Number theory & arithmetic geometry Exact statementFor every integer n≥1 and every number field F, there exist constants δ n,F 0 and A n,F 0 such that |L 1/2,π |≤A n,F C π ^ 1/4−δ n,F for every unitary cuspidal automorphic representation π of GL n A F , where L s,π is the standard finite-part L-function and C π its analytic conductor. The bound must be uniform simultaneously in all finite and archimedean parameters of π; imaginary unitary twists are included. Any fixed positive exponent saving suffices; the optimal exponent is not requested. Why it mattersA proof would give a uniform power-saving bound for central automorphic L-values at every fixed rank and number field, even when all arithmetic and analytic parameters vary together. A disproof would rule out any such saving in at least one of these families and contradict the corresponding generalized Lindelöf prediction. Status Open · subcases solved Status reviewed Sources - Bounds for standard L-functions https://ultronozm.github.io/math/standard.html Statement source Problem familyUniform Subconvexity for Standard Automorphic L-Functions on GL n · 1 problem in this edition. Family membership alone does not assert an implication. - 146 · Uniform Subconvexity for Standard Automorphic L-Functions on GL n this problem Ranking uncertainty90% source-fit bootstrap rank range: 128–171. Source-fit opponent count: 25. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 146. Rank 147 Log-Rank Conjecture in Communication Complexity Two people each hold one input to a Boolean function and exchange bits to compute its value. Is their minimum deterministic communication always bounded by a polynomial in the logarithm of the real rank of the function's ±1 input-output matrix?Cryptography, coding, information & optimization Exact statementProve that there is a universal constant C = 1 such that, for every Boolean function f:X x Y - {-1,1} on finite nonempty sets X,Y, D f <= C log 2 max{2,rank R M f } ^C, where M f x,y =f x,y is its sign communication matrix and D f is deterministic two-party communication complexity in bits. Wording clarified after assessmentEarlier comparisons used the wording below. A recorded review treats the published wording above as a clarification of the same intended problem. The wording review itself adds no comparison judgments. Earlier assessed wording Prove that there is a universal constant C such that every Boolean function f has deterministic two-party communication complexity D f at most C log rank R M f ^C. Wording review recorded . Why it mattersA proof would turn an algebraic property of the function's matrix into a bound on the information the parties must exchange. The bound would grow polynomially in logarithmic rank while allowing constant communication at small ranks. Status Open Status review date not recorded in this edition Sources Problem familyLog-Rank Conjecture in Communication Complexity · 1 problem in this edition. Family membership alone does not assert an implication. - 147 · Log-Rank Conjecture in Communication Complexity this problem Ranking uncertainty90% source-fit bootstrap rank range: 113–186. Source-fit opponent count: 20. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 147. Rank 148 Global Regularity of Solutions to the Inviscid Surface Quasi-Geostrophic Equation In the inviscid surface quasi-geostrophic model, a quantity transported across the plane also determines the velocity that carries it. Must every sufficiently smooth, decaying, finite-energy initial state remain smooth for all time?Analysis & PDE Exact statementFor the inviscid SQG equation partial t theta + u dot grad theta = 0 on R^2, with u = R-perp theta, prove global smooth existence for every sufficiently smooth decaying finite-energy initial datum, or construct such a datum whose solution develops a finite-time singularity. Why it mattersA proof would exclude finite-time singularities throughout the specified class despite the absence of dissipation. A blow-up example would show that this scalar transport law can generate a singularity from smooth initial data. Status Open Status review date not recorded in this edition Sources Problem familyGlobal Regularity of Solutions to the Inviscid Surface Quasi-Geostrophic Equation · 1 problem in this edition. Family membership alone does not assert an implication. - 148 · Global Regularity of Solutions to the Inviscid Surface Quasi-Geostrophic Equation this problem Ranking uncertainty90% source-fit bootstrap rank range: 123–183. Source-fit opponent count: 36. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 148. Rank 149 Murre's Chow-Kunneth Conjectures Does every smooth projective complex variety admit a Chow–Künneth splitting for rational algebraic cycles into cohomological degrees, with every such splitting placing codimension-j cycles only in degrees j through 2j and producing one intrinsic filtration whose first step consists exactly of cycles invisible in cohomology?Number theory & arithmetic geometry Exact statementFor every smooth projective irreducible complex variety X of dimension d, let CH^j X Q denote cycles modulo rational equivalence, with Betti Q-cohomology. A correspondence pi in CH^d X x X Q acts by pi a = p 2 p 1^ a .pi . Murre's full A-D package asserts: A there exists a Chow-Kunneth decomposition, namely mutually orthogonal composition-idempotents pi 0,...,pi 2d in CH^d X x X Q summing to the diagonal and inducing projection onto H^i X,Q for pi i; B for EVERY such decomposition and 0<=j<=d, pi i acts as zero on CH^j when i<j or i 2j; C for EVERY such decomposition define F^r CH^j as the intersection of ker pi i for max 0,2j-r+1 <=i<=2j, r =0, with empty intersection F^0=CH^j; these filtrations are equal for ALL decompositions satisfying A; D for EVERY such decomposition, F^1 CH^j equals ker CH^j X Q - H^ 2j X,Q . Under A and B, F^ j+1 =0. No multiplicative or self-dual decomposition, explicit construction algorithm, functorial-extra requirement, or arbitrary-base-field enlargement is included. Why it mattersThe full package would organize rational algebraic cycles into a filtration independent of the chosen projectors. Its first step would capture exactly the cycles invisible to Betti cohomology, with the prescribed vanishing properties throughout. Status Open · subcases solved Status review date not recorded in this edition Sources Problem familyMurre's Chow-Kunneth Conjectures · 1 problem in this edition. Family membership alone does not assert an implication. - 149 · Murre's Chow-Kunneth Conjectures this problem Ranking uncertainty90% source-fit bootstrap rank range: 126–172. Source-fit opponent count: 23. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 149. Rank 150 Congruent Number Problem Exactly which positive integers occur as the area of a right triangle whose three side lengths are rational numbers?Number theory & arithmetic geometry Elliptic curve ranks and congruent numbers · 3 problems family-context-problem-congruent-number-problem Exact statementDetermine exactly which positive integers are congruent numbers, i.e. areas of right triangles with all three sides rational, equivalently for which squarefree n the elliptic curve y^2 = x^3 - n^2 x has a rational point with y = 0. Why it mattersA complete answer would classify these rational triangles by their integer areas. For squarefree n, it would also determine exactly when y² = x³ − n²x has a rational point with nonzero y. Status Open · subcases solved Status review date not recorded in this edition Sources Problem familyElliptic curve ranks and congruent numbers · 3 problems in this edition. Family membership alone does not assert an implication. - 5 · Birch and Swinnerton-Dyer Conjecture problem-birch-and-swinnerton-dyer-conjecture - 150 · Congruent Number Problem this problem - 194 · Parity conjecture for elliptic curves problem-parity-conjecture-for-elliptic-curves Ranking uncertainty90% source-fit bootstrap rank range: 115–203. Source-fit opponent count: 24. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 150. Rank 151 Order of Magnitude of the Largest Three-Term-Progression-Free Sets As N grows, what is the asymptotic order of the largest subset of {1,…,N} containing no three distinct numbers in arithmetic progression?Combinatorics & discrete geometry Exact statementLet r 3 N be the largest cardinality of a subset of {1,...,N} containing no three distinct terms in arithmetic progression. Determine the asymptotic order of r 3 N as N tends to infinity. Why it mattersA matching asymptotic estimate would determine how large an integer set can be while avoiding the simplest nontrivial arithmetic progression. It would identify the density scale at which this additive pattern becomes unavoidable. Status Open Open. Bloom--Sisask 2023 and Raghavan 2026 improve upper bounds for progression-free sets but leave the matching asymptotic-order problem unresolved. Status review date not recorded in this edition Sources - Bloom and Sisask, An improvement to the Kelley-Meka bounds on three-term arithmetic progressions 2023 https://arxiv.org/html/2309.02353 Statement source - Status source for Order of Magnitude of the Largest Three-Term-Progression-Free Sets https://arxiv.org/html/2309.02353 Status source - Status source for Order of Magnitude of the Largest Three-Term-Progression-Free Sets https://arxiv.org/abs/2603.27045 Status source Problem familyOrder of Magnitude of the Largest Three-Term-Progression-Free Sets · 1 problem in this edition. Family membership alone does not assert an implication. - 151 · Order of Magnitude of the Largest Three-Term-Progression-Free Sets this problem Ranking uncertainty90% source-fit bootstrap rank range: 128–182. Source-fit opponent count: 24. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 151. Rank 152 Consistency of Reinhardt Cardinals without Choice Can ZF set theory, without the axiom of choice, consistently admit a nonidentity self-embedding of the entire set-theoretic universe that preserves statements in the original language of set theory, with the embedding included in the expanded theory's language?Logic, foundations & set theory Exact statementDetermine whether ZFR is consistent, where ZFR is the first-order theory ZF j plus the assertion that a nontrivial function symbol j:V- V is Sigma 1-elementary and hence fully elementary as a theorem scheme. Why it mattersThe answer would determine whether this form of self-embedding can coexist with the specified set-theoretic axioms. An inconsistency proof would rule it out even when the axiom of choice is absent. Status Open Status review date not recorded in this edition Sources Problem familyConsistency of Reinhardt Cardinals without Choice · 1 problem in this edition. Family membership alone does not assert an implication. - 152 · Consistency of Reinhardt Cardinals without Choice this problem Ranking uncertainty90% source-fit bootstrap rank range: 112–215. Source-fit opponent count: 24. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 152. Rank 153 Asymptotic tightness of the binary Gilbert–Varshamov bound Binary codes separate bit strings so errors can be detected or corrected. For every fixed relative separation δ between zero and one-half, is the best asymptotic information rate exactly the Gilbert–Varshamov value 1−H₂ δ , where H₂ is binary entropy, even when nonlinear codes are allowed?Cryptography, coding, information & optimization Exact statementFor integers n =1 and d =0, let A 2 n,d be the maximum cardinality of a binary code C subset {0,1}^n, not necessarily linear, with Hamming distance at least d between distinct codewords. For each fixed real 0<delta<1/2 define R 2 delta =limsup {n- infinity} n^ -1 log 2 A 2 n,floor delta n and H 2 delta =-delta log 2 delta - 1-delta log 2 1-delta . Is R 2 delta =1-H 2 delta for every such delta? Why it mattersEquality would determine the best exponential growth rate of binary code sizes at every such distance. A disproof would show a genuine rate improvement somewhere, without necessarily providing efficiently constructible or decodable codes. Status Open Status review date not recorded in this edition Sources Problem familyAsymptotic tightness of the binary Gilbert–Varshamov bound · 1 problem in this edition. Family membership alone does not assert an implication. - 153 · Asymptotic tightness of the binary Gilbert–Varshamov bound this problem Ranking uncertainty90% source-fit bootstrap rank range: 133–176. Source-fit opponent count: 46. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 153. Rank 154 Colliot-Thelene Conjecture: Brauer-Manin Obstruction Controls Rational Points on Rationally Connected Varieties On smooth proper geometrically rationally connected varieties over number fields, does passing the Brauer–Manin compatibility test ensure that local solutions can be approximated arbitrarily closely by rational points?Number theory & arithmetic geometry Exact statementLet X be a smooth, proper, geometrically irreducible variety over a number field k whose base change to an algebraic closure is rationally connected. Prove that X k is dense in the Brauer-Manin set X A k ^{Br X }; in particular the Brauer-Manin obstruction is the only obstruction to the Hasse principle and weak approximation for such varieties. Why it mattersA proof would make the Brauer–Manin obstruction the complete obstruction to existence and weak approximation of rational points in this class. A counterexample would demonstrate an additional obstacle beyond that test. Status Open · subcases solved Status review date not recorded in this edition Sources Problem familyColliot-Thelene Conjecture: Brauer-Manin Obstruction Controls Rational Points on Rationally Connected Varieties · 1 problem in this edition. Family membership alone does not assert an implication. - 154 · Colliot-Thelene Conjecture: Brauer-Manin Obstruction Controls Rational Points on Rationally Connected Varieties this problem Ranking uncertainty90% source-fit bootstrap rank range: 124–195. Source-fit opponent count: 21. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 154. Rank 155 Grothendieck's Variational Hodge Conjecture In a smooth projective family over a smooth connected complex algebraic base, consider a cohomology class transported consistently across the fibres by a global algebraic horizontal section. If it represents an algebraic cycle with rational coefficients on one fibre, must it do so on every fibre?Geometry & topology Exact statementFor every smooth projective morphism f:X- S of smooth connected complex algebraic varieties, every integer p =0, and every global algebraic horizontal section xi of the relative algebraic de Rham cohomology bundle R^ 2p f Omega^. X/S for the Gauss-Manin connection, if xi s0 is the de Rham cycle class of a rational algebraic codimension-p cycle on one fiber X s0, then xi s is such a rational algebraic cycle class on every fiber X s for s in S C . Why it mattersA proof would show that algebraicity persists under the specified global flat transport through a family. It would not automatically produce one cycle on the entire family or an effective method for constructing the fibrewise cycles. Status Open · subcases solved Status review date not recorded in this edition Sources Problem familyGrothendieck's Variational Hodge Conjecture · 1 problem in this edition. Family membership alone does not assert an implication. - 155 · Grothendieck's Variational Hodge Conjecture this problem Ranking uncertainty90% source-fit bootstrap rank range: 128–196. Source-fit opponent count: 24. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 155. Rank 156 Density of Hyperbolicity for Rational Maps For each degree at least two, can every rational map of the Riemann sphere be approximated arbitrarily closely by maps whose critical points all approach attracting periodic cycles?Probability, ergodic theory & dynamics Density of hyperbolicity · 2 problems family-context-problem-density-of-hyperbolicity-for-rational-maps Exact statementDetermine whether hyperbolic rational maps, whose critical points are all attracted to attracting periodic cycles, are dense in the space of rational maps of the Riemann sphere of each fixed degree d = 2. Why it mattersA positive answer would make this specific attracting-cycle behavior available arbitrarily close to every map of the same degree. A negative answer would reveal a region of parameter space containing no such maps. Status Open Status review date not recorded in this edition Sources Problem familyDensity of hyperbolicity · 2 problems in this edition. Family membership alone does not assert an implication. - 156 · Density of Hyperbolicity for Rational Maps this problem - 220 · Density of Hyperbolicity for Complex Polynomials problem-density-of-hyperbolicity-for-complex-polynomials Ranking uncertainty90% source-fit bootstrap rank range: 130–192. Source-fit opponent count: 33. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 156. Rank 157 Leopoldt's conjecture For every number field and prime p, do the p-adic logarithms of its independent units retain the full expected number of independent directions, so the p-adic regulator never loses rank?Number theory & arithmetic geometry Exact statementProve or disprove that for every number field K and every prime p, the Leopoldt defect of K at p is zero, equivalently that the p-adic regulator has the conjectured full rank. Why it mattersA proof would rule out unexpected rank loss in these p-adic regulators uniformly across number fields and primes. A counterexample would identify a specific field and prime where the expected independence fails. Status Open · subcases solved Status review date not recorded in this edition Sources Research on ProofAtlas Problem familyLeopoldt's conjecture · 1 problem in this edition. Family membership alone does not assert an implication. - 157 · Leopoldt's conjecture this problem Ranking uncertainty90% source-fit bootstrap rank range: 130–194. Source-fit opponent count: 27. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 157. Rank 158 Erdos-Rado Sunflower Conjecture For every integer r≥3, is there a constant C depending only on r such that any family of more than C^k sets of size at most k contains r sets whose pairwise intersections are all the same?Combinatorics & discrete geometry Exact statementFor every fixed integer r = 3, prove or disprove that there is a constant C r such that every family of more than C r^k sets of size at most k contains an r-petal sunflower. Why it mattersA proof would give a purely exponential threshold forcing a sunflower: sets sharing one common core and otherwise disjoint. It would quantify when sufficiently large set families must contain this reusable combinatorial pattern. Status Open Status review date not recorded in this edition Sources Research on ProofAtlas Problem familyErdos-Rado Sunflower Conjecture · 1 problem in this edition. Family membership alone does not assert an implication. - 158 · Erdos-Rado Sunflower Conjecture this problem Ranking uncertainty90% source-fit bootstrap rank range: 137–193. Source-fit opponent count: 48. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 158. Rank 159 Manins Conjecture on the Asymptotic Density of Rational Points on Fano Varieties A point's height measures its arithmetic size. On every smooth Fano variety over a number field, does counting points up to anticanonical height B on a suitable open part give a leading term cB log B ^ ρ−1 , with ρ the Picard rank?Number theory & arithmetic geometry Exact statementProve that for any smooth Fano variety X over a number field K with ample anticanonical bundle, the number of rational points of anticanonical height <= B on a Zariski open subset satisfies N U, B ~ c B log B ^{rho X -1}. Why it mattersThe formula would connect the growth of rational-point counts to geometric invariants and the chosen height. It would predict a leading asymptotic on the selected open subset as the height cutoff increases. Status Open Status review date not recorded in this edition Sources Problem familyManins Conjecture on the Asymptotic Density of Rational Points on Fano Varieties · 1 problem in this edition. Family membership alone does not assert an implication. - 159 · Manins Conjecture on the Asymptotic Density of Rational Points on Fano Varieties this problem Ranking uncertainty90% source-fit bootstrap rank range: 139–196. Source-fit opponent count: 33. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 159. Rank 160 Tate's rational leading-term Stark conjecture For finite Galois extensions of number fields, do the ratios linking unit regulators to the first nonzero terms of Artin L-functions always lie in their corresponding character fields, compatibly with every automorphism of the complex number field?Number theory & arithmetic geometry Explicit class field theory and Stark conjectures · 3 problems family-context-problem-tates-rational-leading-term-stark-conjecture Exact statementFor every finite Galois extension L/K of number fields with group G, let S be any finite set of places of K containing the archimedean and ramified places. Let X be the augmentation kernel of the free Q-module on places of L above S, U=Q tensor Z O L,S ^ , and f:X- U any Q G -isomorphism. Define lambda:R tensor U- R tensor X by lambda u =-sum w log|u| w w, using product-formula normalized absolute values. For each irreducible complex character chi of G, set R S chi,f =det C lambda f acting by postcomposition on Hom C G V dual chi ,C tensor X and A S chi,f =R S chi,f /L S^ 0,chi , where L S^ is the first nonzero Laurent coefficient of the S-truncated Artin L-function and an empty determinant is1. Is A S chi,f in Q chi , with sigma A S chi,f =A S chi^sigma,f for every sigma in Aut C ? Why it mattersA proof would give these analytically defined ratios precise algebraic locations and Galois symmetries. It would establish the rational leading-term relation without automatically proving integral refinements or constructing Stark units effectively. Status Open · subcases solved Status review date not recorded in this edition Sources Research on ProofAtlas Problem familyExplicit class field theory and Stark conjectures · 3 problems in this edition. Family membership alone does not assert an implication. - 36 · Extension of Kroneckers Theorem on Abelian Fields to Any Algebraic Base Field problem-extension-of-kroneckers-theorem-on-abelian-fields-to-any-algebraic-base-field - 160 · Tate's rational leading-term Stark conjecture this problem - 197 · Rank One Abelian Stark Conjecture problem-rank-one-abelian-stark-conjecture Ranking uncertainty90% source-fit bootstrap rank range: 139–193. Source-fit opponent count: 21. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 160. Rank 161 BPL = L Derandomizing Bounded-Error Logspace Can randomness always be removed from a polynomial-time decision algorithm that uses only logarithmic working memory and has error probability at most one-third, without increasing its memory beyond a logarithmic amount?Cryptography, coding, information & optimization Derandomizing logarithmic space · 2 problems family-context-problem-bpl-l-derandomizing-bounded-error-logspace Exact statementProve or disprove that every language decidable by a probabilistic Turing machine using O log n workspace, polynomial time, and bounded two-sided error is decidable by a deterministic Turing machine using O log n workspace. Why it mattersEquality would remove randomness from every computation in this restricted-memory class without increasing the asymptotic space bound. A separation would show that randomness increases what logarithmic memory can decide. Status Open Status review date not recorded in this edition Sources Problem familyDerandomizing logarithmic space · 2 problems in this edition. Family membership alone does not assert an implication. - 161 · BPL = L Derandomizing Bounded-Error Logspace this problem - 260 · RL versus L problem-rl-versus-l Ranking uncertainty90% source-fit bootstrap rank range: 135–197. Source-fit opponent count: 32. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 161. Rank 162 NP versus the Existential Theory of the Reals If a system of polynomial equations and inequalities with integer coefficients has a real solution, must there be a certificate of polynomial size that an ordinary computer can check in polynomial time?Applied & computational mathematics Exact statementDetermine whether every problem decidable in the existential theory of the reals exists R can be decided in nondeterministic polynomial time, i.e., decide whether NP = exists R. Why it mattersEquality would give polynomial-size, efficiently checkable certificates for every yes-instance in this real-feasibility complexity class. A separation would establish that at least one such problem lies beyond NP. Status Open Status review date not recorded in this edition Sources Problem familyNP versus the Existential Theory of the Reals · 1 problem in this edition. Family membership alone does not assert an implication. - 162 · NP versus the Existential Theory of the Reals this problem Ranking uncertainty90% source-fit bootstrap rank range: 130–204. Source-fit opponent count: 27. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 162. Rank 163 Cramer Conjecture on the Limsup of Normalized Prime Gaps Are the largest long-run gaps between consecutive primes on the scale of the square of the logarithm of the smaller prime, with the limiting upper ratio exactly one?Number theory & arithmetic geometry Prime gaps · 2 problems family-context-problem-cramer-conjecture-prime-gaps Exact statementProve that lim sup {n - infinity} p {n+1} - p n / log p n ^2 = 1, where p n is the n-th prime. Why it mattersThe assertion fixes an exact asymptotic constant for the largest normalized gaps, beyond a bound of the same order. It requires gaps approaching this normalized size arbitrarily far out while excluding any larger limiting upper value. Status Open Status review date not recorded in this edition Sources Research on ProofAtlas Problem familyPrime gaps · 2 problems in this edition. Family membership alone does not assert an implication. - 163 · Cramer Conjecture on the Limsup of Normalized Prime Gaps this problem - 323 · Legendre's Conjecture on Primes Between Consecutive Squares problem-legendres-conjecture Ranking uncertainty90% source-fit bootstrap rank range: 142–195. Source-fit opponent count: 45. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 163. Rank 164 Berman-Hartmanis Isomorphism Conjecture Are all NP-complete decision problems equivalent by a reversible relabeling of binary-string inputs, where both the relabeling and its inverse are computable in polynomial time and the relabeling preserves both yes and no answers?Theoretical computer science Exact statementProve or disprove that for every two languages A and B complete for NP under polynomial-time many-one reductions, there is a bijection h on binary strings such that h A =B and both h and its inverse are computable in polynomial time. Why it mattersA proof would show that all such complete languages have the same structure up to efficient reversible relabeling. A counterexample would distinguish two complete languages despite their sharing the same reduction-based completeness property. Status Open Status review date not recorded in this edition Sources Problem familyBerman-Hartmanis Isomorphism Conjecture · 1 problem in this edition. Family membership alone does not assert an implication. - 164 · Berman-Hartmanis Isomorphism Conjecture this problem Ranking uncertainty90% source-fit bootstrap rank range: 122–232. Source-fit opponent count: 20. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 164. Rank 165 Smooth Four-Dimensional Schoenflies Problem Does every smoothly embedded three-sphere in the four-sphere cut it into two pieces whose closures are each smoothly equivalent to the standard four-dimensional ball?Geometry & topology Smooth four-dimensional spheres · 2 problems family-context-problem-smooth-four-dimensional-schoenflies-problem Exact statementDetermine whether every smooth embedding of S^3 in S^4 is smoothly unknotted; equivalently, whether each of its two complementary closures is diffeomorphic to the standard 4-ball B^4. Why it mattersA positive answer would rule out smooth knotting of three-spheres in the four-sphere. A negative answer would reveal an embedding whose complementary closures are not both smoothly equivalent to standard balls. Status Open · subcases solved Status review date not recorded in this edition Sources Research on ProofAtlas Problem familySmooth four-dimensional spheres · 2 problems in this edition. Family membership alone does not assert an implication. - 21 · Smooth four-dimensional Poincare conjecture problem-smooth-four-dimensional-poincare-conjecture - 165 · Smooth Four-Dimensional Schoenflies Problem this problem Ranking uncertainty90% source-fit bootstrap rank range: 139–201. Source-fit opponent count: 29. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 165. Rank 166 Uniform Boundedness Conjecture for Rational Points on Curves For each genus of at least two and each number field, is there one finite bound on rational points that works for every smooth curve of that genus over that field?Number theory & arithmetic geometry Exact statementFor each integer g = 2 and number field F, prove that there is a constant c g,F such that every smooth curve C of genus g over F has at most c g,F F-rational points. Why it mattersA proof would make the number of rational points uniformly controllable while the curve varies. A disproof would produce arbitrarily large point counts even after both genus and number field are fixed. Status Open · subcases solved Status review date not recorded in this edition Sources Problem familyUniform Boundedness Conjecture for Rational Points on Curves · 1 problem in this edition. Family membership alone does not assert an implication. - 166 · Uniform Boundedness Conjecture for Rational Points on Curves this problem Ranking uncertainty90% source-fit bootstrap rank range: 134–207. Source-fit opponent count: 34. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 166. Rank 167 Keating-Snaith Conjecture on Moments of the Riemann Zeta Function For every real k 0, does the average of |ζ 1/2+it |^ 2k over 0≤t≤T grow like log T ^ k² times the exact Keating–Snaith constant, giving the predicted statistics of zeta's size on the critical line?Number theory & arithmetic geometry Exact statementFor every fixed real k 0, let I k T =integral 0^T |zeta 1/2+it |^ 2k dt. Prove that I k T is asymptotic to a k G 1+k ^2/G 1+2k times T log T ^ k^2 , where G is the Barnes G-function and a k=product p 1-1/p ^ k^2 sum m =0 Gamma m+k / Gamma k m ^2 p^ -m . Why it mattersThe formula would determine both the leading growth and the exact leading constant for every positive moment on the critical line. These moments quantify different aspects of the zeta function's size distribution. Status Open · subcases solved October 6, 2026: De Faveri and Pandey 1 October improve the sixth-moment upper bound to O T^ 37/30+epsilon . This is not the conjectured asymptotic, even for the sixth moment, and does not resolve all fixed real moments. The complete proof was not audited in this review. Status reviewed Sources Problem familyKeating-Snaith Conjecture on Moments of the Riemann Zeta Function · 1 problem in this edition. Family membership alone does not assert an implication. - 167 · Keating-Snaith Conjecture on Moments of the Riemann Zeta Function this problem Ranking uncertainty90% source-fit bootstrap rank range: 138–217. Source-fit opponent count: 20. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 167. Rank 168 Voevodsky's Smash-Nilpotence Conjecture If an algebraic cycle with rational coefficients has zero intersection number against every complementary cycle, must multiplying copies of it on a product of sufficiently many copies of the variety make it zero modulo rational equivalence, for every smooth projective variety over any field?Number theory & arithmetic geometry Exact statementFor every field k, every smooth projective k-variety X of dimension d, every integer 0<=r<=d, and every class alpha in CH^r X Q codimension-r cycles modulo rational equivalence with rational coefficients , assume deg alpha.beta =0 for every beta in CH^ d-r X Q. Voevodsky's smash-nilpotence assertion is that there exists an integer N =1, depending on alpha, for which the N-fold EXTERNAL product alpha times ... times alpha is zero in CH^ Nr X^N Q, where X^N is the fibre product over k. No algebraic-closure, characteristic-zero or geometric-connectedness hypothesis is imposed; disconnected smooth projective schemes are interpreted on their pure-dimensional components. The reverse implication, smash-nilpotent implies numerically trivial, is known, so this is equality of numerical and smash equivalence for rational cycles. Neither intersection powers on X, composition powers of correspondences, integral coefficients, nor a uniform or computable exponent replaces this assertion. Why it mattersA proof would characterize numerical triviality by actual vanishing after enough external products. The required exponent may depend on the cycle, and the products take place on powers of the variety rather than within the original variety. Status Open · subcases solved Status review date not recorded in this edition Sources Problem familyVoevodsky's Smash-Nilpotence Conjecture · 1 problem in this edition. Family membership alone does not assert an implication. - 168 · Voevodsky's Smash-Nilpotence Conjecture this problem Ranking uncertainty90% source-fit bootstrap rank range: 136–216. Source-fit opponent count: 21. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 168. Rank 169 The Free Group Factor Isomorphism Problem Can the von Neumann algebras associated with free groups on different numbers of generators be isomorphic, in particular those associated with two and three generators?Algebra, representation & category theory Exact statementDetermine whether L F m and L F n are isomorphic when m = n, especially whether L F 2 is isomorphic to L F 3 . Why it mattersThe answer would determine whether these operator algebras retain the number of free generators or can lose that distinction. It tests what information survives the passage from a free group to its associated factor. Status Open Status review date not recorded in this edition Sources Problem familyThe Free Group Factor Isomorphism Problem · 1 problem in this edition. Family membership alone does not assert an implication. - 169 · The Free Group Factor Isomorphism Problem this problem Ranking uncertainty90% source-fit bootstrap rank range: 149–198. Source-fit opponent count: 44. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 169. Rank 170 Simplicity of the Nontrivial Zeros of the Riemann Zeta Function Does every nontrivial zero of the Riemann zeta function have multiplicity exactly one, so that the function's derivative is nonzero at each such zero?Number theory & arithmetic geometry Exact statementProve that zeta' rho is nonzero for every nontrivial zero rho of the Riemann zeta function; equivalently, every nontrivial zero of zeta s has multiplicity exactly one. Why it mattersA proof would rule out repeated nontrivial zeros and fix their local multiplicity structure. This addresses how zeta vanishes at its zeros without asserting where those zeros lie. Status Open · subcases solved Status review date not recorded in this edition Sources Problem familySimplicity of the Nontrivial Zeros of the Riemann Zeta Function · 1 problem in this edition. Family membership alone does not assert an implication. - 170 · Simplicity of the Nontrivial Zeros of the Riemann Zeta Function this problem Ranking uncertainty90% source-fit bootstrap rank range: 142–210. Source-fit opponent count: 20. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 170. Rank 171 Weinstein Conjecture on periodic orbits of Reeb flows A contact form determines a distinguished flow called its Reeb flow. On every compact contact manifold without boundary, must every such flow have a trajectory that returns to its starting point?Geometry & topology Exact statementProve that every closed contact manifold admits a periodic orbit of the Reeb flow of any contact form, in all dimensions. Why it mattersA proof would guarantee recurring motion for every flow in this geometric class. A counterexample would give a contact form whose Reeb dynamics avoid periodic trajectories entirely on a closed manifold. Status Open · subcases solved Status review date not recorded in this edition Sources Research on ProofAtlas Problem familyWeinstein Conjecture on periodic orbits of Reeb flows · 1 problem in this edition. Family membership alone does not assert an implication. - 171 · Weinstein Conjecture on periodic orbits of Reeb flows this problem Ranking uncertainty90% source-fit bootstrap rank range: 143–219. Source-fit opponent count: 21. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 171. Rank 172 Separation of P R from NP R in the Blum–Shub–Smale Model In a computer model that performs exact real arithmetic at unit cost, is there a decision problem whose real-number certificates can be checked in polynomial time but which no deterministic polynomial-time machine can solve?Applied & computational mathematics Exact statementIn the standard ordered-real Blum–Shub–Smale model, machines use finitely many built-in real constants, real-valued registers, exact field operations +,-,multiplication,division where defined and order/equality tests at unit cost. Inputs are finite real vectors and size is vector length. Let P R be decision problems solved by deterministic machines in polynomially many steps, and NP R those with polynomial-length real witnesses verifiable in P R. The assertion is P R = NP R. Why it mattersA separation would distinguish verification from computation when exact real operations have unit cost. The conclusion belongs to this real-number machine model and does not automatically establish the corresponding separation for binary-string computation. Status Open Status review date not recorded in this edition Sources Problem familySeparation of P R from NP R in the Blum–Shub–Smale Model · 1 problem in this edition. Family membership alone does not assert an implication. - 172 · Separation of P R from NP R in the Blum–Shub–Smale Model this problem Ranking uncertainty90% source-fit bootstrap rank range: 148–208. Source-fit opponent count: 30. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 172. Rank 173 Conformal Invariance of the Critical Random-Cluster Model with q <= 4 As a critical random-cluster model with cluster weight q≤4 is viewed at finer and finer lattice scales, does its separating interface approach the predicted Schramm–Loewner random curve, whose law is unchanged by conformal changes of coordinates?Probability, ergodic theory & dynamics Exact statementProve that the exploration path of the critical random-cluster FK model with q <= 4 converges in distribution to a Schramm-Loewner evolution SLE {kappa q } as the mesh size goes to zero, i.e., the scaling limit is conformally invariant. Why it mattersA proof would identify the continuum distribution of the exploration path and establish its conformal invariance. It would connect the discrete lattice model to a precisely specified random-curve law in the scaling limit. Status Open · subcases solved Status review date not recorded in this edition Sources Problem familyConformal Invariance of the Critical Random-Cluster Model with q <= 4 · 1 problem in this edition. Family membership alone does not assert an implication. - 173 · Conformal Invariance of the Critical Random-Cluster Model with q <= 4 this problem Ranking uncertainty90% source-fit bootstrap rank range: 118–280. Source-fit opponent count: 21. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 173. Rank 174 Dirichlet Divisor Problem After subtracting the standard main term from the total number of divisors of the integers up to x, what is the optimal power of x that bounds the error as x grows, allowing an arbitrarily small extra exponent?Number theory & arithmetic geometry Lattice point and divisor error bounds · 2 problems family-context-problem-dirichlet-divisor-problem Open · disputed claim Exact statementLet d n count the positive divisors of n and Delta x =sum {n<=x} d n -x log x- 2 gamma-1 x for real x =2. Determine the optimal exponent alpha 2=inf{a: for every epsilon 0, Delta x =O epsilon x^ a+epsilon as x tends to infinity}. Why it mattersThis would identify the sharp power-law threshold for fluctuations in the cumulative divisor count. The target concerns the exponent, rather than finer logarithmic factors in the error. Status Open · disputed claim Open with unverified claims recorded. Mo 2011 and Jabbarov May 2026 revision, Theorem 3.1 claim the quarter-exponent bound for the same unsmoothed error term. No independent validation is established here. Bordelles--Daval January 2026 and Lamzouri May 2026, after the newer claim continue to treat that bound as conjectural. No proof claim is certified or adopted as a theorem. Status review date not recorded in this edition Sources - Karak and Mahatab, The Piltz divisor Problem in Number Fields Using The Resonance Method 2025; journal 2026 https://arxiv.org/html/2506.22587v1 Statement source - Status source for Dirichlet Divisor Problem https://arxiv.org/html/2506.22587v1 Status source - Status source for Dirichlet Divisor Problem https://arxiv.org/html/2601.01905v2 Status source - Status source for Dirichlet Divisor Problem https://arxiv.org/pdf/1105.6155 Status source - Status source for Dirichlet Divisor Problem https://arxiv.org/html/2604.18624v2 Status source - Status source for Dirichlet Divisor Problem https://arxiv.org/html/2605.21476v1 Status source Problem familyLattice point and divisor error bounds · 2 problems in this edition. Family membership alone does not assert an implication. - 174 · Dirichlet Divisor Problem this problem - 262 · Gauss Circle Problem problem-gauss-circle-problem Ranking uncertainty90% source-fit bootstrap rank range: 143–213. Source-fit opponent count: 23. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 174. Rank 175 Capacity of the General Discrete Memoryless Relay Channel What is the highest reliable communication rate through a memoryless channel when a relay can help using only what it has already observed? Find a computable exact answer for every finite-alphabet channel, with reliability measured by average error tending to zero.Cryptography, coding, information & optimizationOpen · disputed claim Exact statementFind a computable exact average-error Shannon-capacity characterization for every finite-alphabet discrete memoryless relay channel with law p y,y r|x,x r , where the relay input at time i depends only on its observations before time i. Why it mattersA computable characterization would determine the reliable communication limit for every channel in this model. A proof that no such characterization exists would instead establish a limit on computably describing those capacities. Status Open · disputed claim The general relay-channel capacity problem remains open. Ponniah’s proposed formula first posted September 14, 2026 is challenged by Shiu’s September 16 counterexample: a strictly causal finite-alphabet channel achieves approximately 1.18872 bits/use, above the formula’s bound of approximately 1.04718. The inspected September 20 revision retains the formula and does not establish a resolution of that objection. Shiu credits ChatGPT-6 Astra with obtaining the counterexample. This refutes the proposed formula, not the possibility of a general capacity characterization. Status reviewed Sources - Ozyilkan, Carpi, Garg and Erkip, Learning-Based Compress-and-Forward Schemes for the Relay Channel https://arxiv.org/html/2405.09534v3 Statement source - Ponniah, On the capacity of the relay channel v2 https://arxiv.org/html/2609.15709v2 Status source · Retrieved Oct 6, 2026 - Shiu, counterexample to the proposed relay-channel capacity formula v1 https://arxiv.org/html/2609.18727v1 Status source · Retrieved Oct 6, 2026 Problem familyCapacity of the General Discrete Memoryless Relay Channel · 1 problem in this edition. Family membership alone does not assert an implication. - 175 · Capacity of the General Discrete Memoryless Relay Channel this problem Ranking uncertainty90% source-fit bootstrap rank range: 150–205. Source-fit opponent count: 24. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 175. Rank 176 Algebraic Independence of e and pi Is it impossible for any nonzero polynomial in two variables with rational coefficients to become zero when its variables are replaced by e and π?Number theory & arithmetic geometry Exponential transcendence · 3 problems family-context-problem-algebraic-independence-of-e-and-pi Exact statementProve that no nonzero polynomial P in Q X,Y satisfies P e,pi =0. Why it mattersA proof would exclude every polynomial relation between these two constants. It would also imply that their sum and product are transcendental, while making no blanket claim about arbitrary exponential expressions involving them. Status Open Status review date not recorded in this edition Sources Problem familyExponential transcendence · 3 problems in this edition. Family membership alone does not assert an implication. - 27 · Schanuel's conjecture problem-schanuel-conjecture - 176 · Algebraic Independence of e and pi this problem - 379 · Four Exponentials Conjecture problem-four-exponentials-conjecture Ranking uncertainty90% source-fit bootstrap rank range: 153–209. Source-fit opponent count: 20. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 176. Rank 177 L-space Conjecture For a closed oriented irreducible rational homology three-sphere, are these three conditions equivalent: its Heegaard Floer homology has more than minimal size; its nontrivial group of loops up to deformation admits a total ordering preserved by left multiplication; and it can be foliated by surfaces with a consistent transverse orientation, each leaf meeting a closed transverse curve?Geometry & topology Exact statementProve or disprove: for a closed, oriented, irreducible rational homology 3-sphere Y, the following are equivalent: 1 Y is not a Heegaard Floer L-space; 2 pi 1 Y is nontrivial and admits a left-invariant strict total order; 3 Y admits a coorientable taut foliation. Wording clarified after assessmentEarlier comparisons used the wording below. A recorded review treats the published wording above as a clarification of the same intended problem. The wording review itself adds no comparison judgments. Earlier assessed wording Prove or disprove: for a closed, oriented, irreducible rational homology 3-sphere Y, the following are equivalent: 1 Y is not a Heegaard Floer L-space its Heegaard Floer homology is not minimal ; 2 pi 1 Y is left-orderable; 3 Y admits a coorientable taut foliation. Wording review recorded . Why it mattersA proof would identify the same class of manifolds through Heegaard Floer homology, an ordering of the fundamental group and a foliation condition. A counterexample would separate at least two of these descriptions. Status Open · subcases solved Status review date not recorded in this edition Sources Problem familyL-space Conjecture · 1 problem in this edition. Family membership alone does not assert an implication. - 177 · L-space Conjecture this problem Ranking uncertainty90% source-fit bootstrap rank range: 131–235. Source-fit opponent count: 20. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 177. Rank 178 Arnold Conjecture on Hamiltonian Fixed Points For Hamiltonian motion on a closed connected symplectic manifold, must its time-one transformation fix at least as many points as the fewest critical points a smooth real-valued function on that manifold can have?Geometry & topology Exact statementProve that every Hamiltonian diffeomorphism of an arbitrary closed connected symplectic manifold has at least Crit M fixed points, where Crit M is the minimum number of critical points of a smooth function on M. Why it mattersA proof would turn a constraint on smooth functions into a universal lower bound on points left fixed by Hamiltonian motion. The minimum allows all smooth functions, so the precise critical-point count matters. Status Open Status review date not recorded in this edition Sources - Formal statement, abstract https://arxiv.org/abs/1609.09192 Statement source Problem familyArnold Conjecture on Hamiltonian Fixed Points · 1 problem in this edition. Family membership alone does not assert an implication. - 178 · Arnold Conjecture on Hamiltonian Fixed Points this problem Ranking uncertainty90% source-fit bootstrap rank range: 138–228. Source-fit opponent count: 21. The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem. 178. Rank 179 Omega n log n Boolean Circuit Lower Bound for Integer Multiplication Does multiplying two n-bit nonnegative integers and producing every bit of their product require at least a constant times n log n arbitrary two-input Boolean gates, however the circuit is arranged, for all sufficiently large n?Theoretical computer science