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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.

read241 min views1 publishedOct 6, 2026

The ProofAtlas Top 500

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 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.

Rank001****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 scienceClassical complexity separations · 5 problems### 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. ### StatusOpen 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 #

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.

Rank002****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 geometryRiemann hypotheses and exceptional zeros · 4 problems### 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. ### StatusOpen 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 #

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.

Rank003****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. ### StatusOpen 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 #

  - [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

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. #

Rank004****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 & topologyHodge conjectures · 3 problems### 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. ### StatusOpen 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 #

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. #

Rank005****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 geometryElliptic curve ranks and congruent numbers · 3 problems### 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. ### StatusOpen 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 #

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. #

Rank007****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 geometryLanglands functoriality and modularity · 2 problems### 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. ### StatusOpen · 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 #

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)

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. #

Rank008****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 & optimizationOne-way functions and permutations · 2 problems### 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. ### StatusOpen 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 #

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. #

Rank009****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 geometryabc and Szpiro conjectures · 2 problems 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. ### StatusOpen · 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 #

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

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. #

Rank010****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 geometryRiemann hypotheses and exceptional zeros · 4 problems### 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. ### StatusOpen 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 #

Problem familyRiemann hypotheses and exceptional zeros · 4 problems in this edition. Family membership alone does not assert an implication.

  • #2 · 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. #

Rank011****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 geometryRiemann hypotheses and exceptional zeros · 4 problems 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. ### StatusOpen · 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 #

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

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. #

Rank012****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. ### StatusOpen 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 #

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. #

Rank013****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 scienceClassical complexity separations · 5 problems### 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. ### StatusOpen 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 #

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. #

Rank014****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 scienceClassical complexity separations · 5 problems### 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. ### StatusOpen 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 #

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. #

Rank015****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. ### StatusOpen · 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 #

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. #

Rank016****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. ### StatusOpen 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 #

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. #

Rank017****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. ### StatusOpen 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 #

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. #

Rank018****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 geometrySimultaneous prime values · 5 problems### 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. ### StatusOpen 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 #

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. #

Rank019****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. ### StatusOpen 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 #

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. #

Rank020****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 geometrySimultaneous prime values · 5 problems### 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. ### StatusOpen 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 #

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.

Rank021****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 & topologySmooth four-dimensional spheres · 2 problems### 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. ### StatusOpen 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. #

Rank022****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 theoryHilbert’s tenth problem · 2 problems### 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. ### StatusOpen 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 #

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

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. #

Rank023****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. ### StatusOpen 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 #

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. #

Rank024****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 geometryHodge conjectures · 3 problems### 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. ### StatusOpen · 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 #

Problem familyHodge conjectures · 3 problems in this edition. Family membership alone does not assert an implication.

  • #4 · Hodge Conjecture
  • #24 · Grothendiecks Generalized Hodge Conjecture (this problem)
  • #90 · Deligne's 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. #

Rank025****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. ### StatusOpen 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 #

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. #

Rank026****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. ### StatusOpen 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 #

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. #

Rank027****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 geometryExponential transcendence · 3 problems### 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. ### StatusOpen Schanuel remains open. Reformulations and almost-everywhere Schanuel properties do not establish its universal algebraic-independence conclusion. Status reviewed ### Sources #

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. #

Rank028****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. ### StatusOpen 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 #

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. #

Rank029****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 scienceUnique Games and approximation thresholds · 2 problems### 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. ### StatusOpen 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

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. #

Rank030****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. ### StatusOpen · 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 #

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. #

Rank031****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. ### StatusOpen 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 #

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. #

Rank032****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 scienceAlgebraic computation lower bounds · 3 problems### 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. ### StatusOpen 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 #

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. #

Rank033****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. ### StatusOpen 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 #

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. #

Rank034****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 scienceClassical complexity separations · 5 problems### 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. ### StatusOpen 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 #

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. #

Rank035****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. ### StatusOpen 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

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. #

Rank036****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 geometryExplicit class field theory and Stark conjectures · 3 problems### 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. ### StatusOpen · 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 #

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. #

Rank037****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. ### StatusOpen · 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 #

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. #

Rank038****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 & PDEKakeya, restriction and smoothing estimates · 4 problems### 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. ### StatusOpen · 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 #

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. #

Rank039****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 geometryLanglands functoriality and modularity · 2 problems### 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. ### StatusOpen · 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 #

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. #

Rank040****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. ### StatusOpen 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 #

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. #

Rank041****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. ### StatusOpen 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 #

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. #

Rank042****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 physics3D Ising · 2 problems### 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. ### StatusOpen 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

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

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. #

Rank043****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 geometryGrothendieck’s standard conjectures · 2 problems### 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. ### StatusOpen · 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 #

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

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. #

Rank044****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. ### StatusOpen 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 #

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. #

Rank045****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 & PDEKakeya, restriction and smoothing estimates · 4 problems### 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. ### StatusOpen · 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

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. #

Rank046****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. ### StatusOpen 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 #

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. #

Rank047****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. ### StatusOpen · 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 #

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. #

Rank048****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 geometryRiemann hypotheses and exceptional zeros · 4 problems### 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. ### StatusOpen 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 #

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. #

Rank049****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. ### StatusOpen 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 #

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. #

Rank050****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. ### StatusReviewed 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 #

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. #

Rank051****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. ### StatusOpen 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 #

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. #

Rank052***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. ### StatusOpen 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 #

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. #

Rank053****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 geometrySimultaneous prime values · 5 problems### 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. ### StatusOpen · 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 #

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. #

Rank054****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. ### StatusOpen 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 #

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. #

Rank055****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. ### StatusOpen · 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

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. #

Rank056****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. ### StatusOpen · 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 #

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. #

Rank057****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 & optimizationOne-way functions and permutations · 2 problems### 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. ### StatusOpen 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 #

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. #

Rank058****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 geometrySimultaneous prime values · 5 problems### 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. ### StatusOpen · 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 #

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

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. #

Rank059****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. ### StatusOpen 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 #

  - [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. #

Rank060****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 theoryInverse Galois over Q · 2 problems### 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. ### StatusOpen · 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 #

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

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. #

Rank061****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. ### StatusOpen 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 #

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. #

Rank062****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 scienceAlgebraic computation lower bounds · 3 problems### 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. ### StatusOpen 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 #

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)
  • #62 · Permanent versus Determinant over C (this problem)
  • #340 · Koiran's Real Tau-Conjecture on Real Roots of Sums of Products of Sparse Polynomials

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. #

Rank063****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. ### StatusOpen 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

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. #

Rank064****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. ### StatusOpen 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 #

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. #

Rank065****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. ### StatusOpen 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

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. #

Rank066****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. ### StatusOpen · 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 #

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. #

Rank067****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. ### StatusOpen · 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 #

  - [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. #

Rank068****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. ### StatusOpen 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 #

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. #

Rank069****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. ### StatusOpen 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 #

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. #

Rank070****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. ### StatusOpen 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 #

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. #

Rank071****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)/((2pii)^(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. ### StatusOpen · 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 #

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. #

Rank072****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 & topologyMinimal model conjectures · 3 problems### 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. ### StatusOpen · 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 #

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. #

Rank073****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. ### StatusOpen 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

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. #

Rank074****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 physics3D Ising · 2 problems### 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. ### StatusOpen 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 #

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. #

Rank075****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. ### StatusOpen 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 #

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. #

Rank076****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. ### StatusOpen 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 #

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. #

Rank077****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. ### StatusOpen 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 #

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. #

Rank078****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 & topologyMinimal model conjectures · 3 problems### 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. ### StatusOpen 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 #

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
  • #78 · Abundance Conjecture for the minimal model program (this problem)
  • #193 · 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. #

Rank079****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. ### StatusOpen 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 #

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. #

Rank080****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. ### StatusOpen 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 #

  - [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. #

Rank081****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 geometrySimultaneous prime values · 5 problems### 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. ### StatusOpen · 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.

Rank082****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. ### StatusOpen 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 #

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. #

Rank083****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. ### StatusOpen · 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 #

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. #

Rank084****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. ### StatusOpen · 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 #

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. #

Rank085****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. ### StatusOpen 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 #

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. #

Rank086****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 geometryCampana specialness and rational points · 1 problem### 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. ### StatusOpen · subcases solved Status reviewed ### Sources #

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. #

Rank087****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. ### StatusOpen 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

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. #

Rank088****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. ### StatusOpen 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 #

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. #

Rank089****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 geometryGrothendieck–Teichmüller theory · 2 problems### 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. ### StatusOpen 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 #

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

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. #

Rank090****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 geometryHodge conjectures · 3 problems### 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. ### StatusOpen · 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. #

Rank091****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. ### StatusOpen 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 #

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. #

Rank092****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. ### StatusOpen 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 #

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. #

Rank093****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 theoryGrothendieck’s standard conjectures · 2 problems### 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. ### StatusOpen · 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

Problem familyGrothendieck’s standard conjectures · 2 problems in this edition. Family membership alone does not assert an implication.

  • #43 · Grothendieck's 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. #

Rank094****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. ### StatusOpen 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 #

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. #

Rank095****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. ### StatusResolution 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 #

  - [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

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. #

Rank096****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. ### StatusOpen · 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 #

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. #

Rank097****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. ### StatusOpen 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. #

Rank098****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. ### StatusOpen 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 #

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. #

Rank099****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. ### StatusOpen 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

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. #

Rank100****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. ### StatusReviewed 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 #

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. #

Rank101****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 geometryabc and Szpiro conjectures · 2 problems### 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. ### StatusOpen · 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
  • #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. #

Rank102****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. ### StatusOpen · 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. #

Rank103****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 scienceClassical complexity separations · 5 problems### 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. ### StatusOpen Status review date not recorded in this edition ### Sources #

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. #

Rank104****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. ### StatusOpen 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 #

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. #

Rank105****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. ### StatusOpen 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. #

Rank106****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. ### StatusOpen 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 #

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. #

Rank107****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 & PDEIncompressible fluid regularity · 2 problems### 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. ### StatusOpen · 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
  • #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. #

Rank108****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. ### StatusOpen 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. #

Rank109****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. ### StatusOpen · 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

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. #

Rank110****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. ### StatusOpen 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. #

Rank111****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. ### StatusOpen · 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. #

Rank112****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. ### StatusOpen 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. #

Rank113****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. ### StatusOpen · 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. #

Rank114****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. ### StatusOpen 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. #

Rank115****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. ### StatusOpen 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. #

Rank116****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 theoryInverse Galois over Q · 2 problems### 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. ### StatusOpen · subcases solved Status reviewed ### Sources #

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
  • #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. #

Rank117****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. ### StatusOpen · 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. #

Rank118****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. ### StatusOpen · 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. #

Rank119****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. ### StatusOpen · 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. #

Rank120****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. ### StatusOpen 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. #

Rank121****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 & computationQuantum verification classes · 2 problems### 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. ### StatusOpen · 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. #

Rank122****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 (2pi/N) log|J_{K,N}(exp(2pi*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. ### StatusOpen · 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. #

Rank123****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. ### StatusOpen · 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. #

Rank124****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. ### StatusOpen · 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. #

Rank125****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 geometryZeta zeros and large Dirichlet polynomial values · 2 problems### 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. ### StatusOpen 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

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. #

Rank126****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. ### StatusOpen 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. #

Rank127****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 & dynamicsMandelbrot set boundary · 2 problems### 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. ### StatusOpen · 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. #

Rank128****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. ### StatusOpen Status review date not recorded in this edition ### Sources #

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. #

Rank129****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. ### StatusOpen 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. #

Rank130****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. ### StatusOpen · 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. #

Rank131****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 scienceOptimal propositional proof systems · 2 problems### 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. ### StatusOpen Status reviewed ### Sources #

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

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. #

Rank132****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. ### StatusOpen Status review date not recorded in this edition ### Sources #

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. #

Rank133****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. ### StatusOpen 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. #

Rank134****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. ### StatusOpen · 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. #

Rank135****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. ### StatusOpen 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. #

Rank136****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. ### StatusOpen · 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. #

Rank137****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. ### StatusOpen Status review date not recorded in this edition ### Sources #

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. #

Rank138****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 theoryGroup approximation and determinants · 2 problems### 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. ### StatusOpen 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. #

Rank139****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. ### StatusOpen · subcases solved Status review date not recorded in this edition ### Sources #

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. #

Rank140****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. ### StatusOpen · 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. #

Rank141****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. ### StatusOpen · 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. #

Rank142****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. ### StatusOpen · subcases solved Status review date not recorded in this edition ### Sources #

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. #

Rank143****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. ### StatusOpen Status review date not recorded in this edition ### Sources #

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. #

Rank144****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. ### StatusOpen 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. #

Rank145****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. ### StatusOpen · 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. #

Rank146****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. ### StatusOpen · subcases solved Status reviewed ### Sources #

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. #

Rank147****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. ### StatusOpen 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. #

Rank148****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. ### StatusOpen 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. #

Rank149****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. ### StatusOpen · 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. #

Rank150****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 geometryElliptic curve ranks and congruent numbers · 3 problems### 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. ### StatusOpen · 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. #

Rank151****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. ### StatusOpen 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. #

Rank152****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. ### StatusOpen 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. #

Rank153****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. ### StatusOpen 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. #

Rank154****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. ### StatusOpen · 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. #

Rank155****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. ### StatusOpen · 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. #

Rank156****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 & dynamicsDensity of hyperbolicity · 2 problems### 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. ### StatusOpen 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

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. #

Rank157****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. ### StatusOpen · 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. #

Rank158****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. ### StatusOpen 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. #

Rank159****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. ### StatusOpen 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. #

Rank160****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 geometryExplicit class field theory and Stark conjectures · 3 problems### 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. ### StatusOpen · 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. #

Rank161****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 & optimizationDerandomizing logarithmic space · 2 problems### 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. ### StatusOpen 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. #

Rank162****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. ### StatusOpen 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. #

Rank163****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 geometryPrime gaps · 2 problems### 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. ### StatusOpen 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

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. #

Rank164****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. ### StatusOpen 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. #

Rank165****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 & topologySmooth four-dimensional spheres · 2 problems### 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. ### StatusOpen · 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. #

Rank166****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. ### StatusOpen · 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. #

Rank167****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. ### StatusOpen · 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. #

Rank168****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. ### StatusOpen · 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. #

Rank169****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. ### StatusOpen 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. #

Rank170****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. ### StatusOpen · 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. #

Rank171****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. ### StatusOpen · 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. #

Rank172****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. ### StatusOpen 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. #

Rank173****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. ### StatusOpen · 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. #

Rank174****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 geometryLattice point and divisor error bounds · 2 problems 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. ### StatusOpen · 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. #

Rank175****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. ### StatusOpen · 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. #

Rank176****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 geometryExponential transcendence · 3 problems### 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. ### StatusOpen 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
  • #176 · Algebraic Independence of e and pi (this problem)
  • #379 · 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. #

Rank177****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. ### StatusOpen · 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. #

Rank178****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. ### StatusOpen Status review date not recorded in this edition ### Sources #

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. #

Rank179****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 #

source & further reading
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