{"slug": "more-small-solutions-to-a-new-infinite-family-a-4-b-4-c-4-d-4-a-27b-27c-27d-4", "title": "More small solutions to a new infinite family $a^4+b^4+c^4+d^4 = (a+27b+27c+27d)^4$?", "summary": "Mathematicians Matej Veselovac and Tito Piezas III used Anthropic's Claude AI to discover a new infinite family of solutions to the equation a^4+b^4+c^4+d^4 = (a+27b+27c+27d)^4, building on a database of over 45,000 solutions compiled by Eugene Go. The AI identified 11 'interesting' linear relations among terms, including one where (a+e)/(b+c+d) equals a small rational cube such as 3^3, leading to the new family. The work reduces to an elliptic curve with positive rank, suggesting further solutions exist.", "body_md": "**Q:** We are looking for rational $(D,r)$,\n\n$$D^2=-(18958669594580211381729967107 + 1093490321304049798772416r + 15172317493269316128r^2 + 3306770731944r^3 + 35534992r^4)$$\n\nwith one solution as $r = −48044056139/1242748$ derived from a \"seed\". However, we hope there may be $r_n$ with * smaller* height. The above is birationally equivalent to an elliptic curve with positive rank, so there should be more $r_n$. Any \"small\" point will lead to another solution to the equation in the title. (I don't have Magma or PARI's \"hyperellratpoints\", only Mathematica.) Details below.\n\n**I. The first family**\n\nBack in the pre-AI years of 2008, Jacobi-Madden found the infinite family,\n\n$$a^4+b^4+c^4+d^4 = (a+b+c+d)^4$$\n\nJust a few days ago, **Matej Veselovac** in an [MSE post](https://math.stackexchange.com/a/5146910/4781) found the **new** family,\n\n$$a^4+b^4+c^4+d^4 = (a+27b+27c+27d)^4$$\n\nwhere $k^3$ is defined as the coefficient of $(b,c,d)$ on the RHS. A total of **nine** $k^3$ for some small rational $k$ were found. Jacobi had to peruse only the first **30** solutions of the $(4,1,4)$ to find his first example. But since Matej had to wade through more than **45,000** solutions of [Eugene Go's database](https://drive.google.com/file/d/1lEEdTOVH_7V-dHQl1PjqPx5Emw8K4qm5/view?usp=drive_link) to find patterns, it seems **AI** like Anthropic's Claude Fable proves useful when ascertaining which relations are interesting, and which are random.\n\n**II. The Story**\n\nMatej and I met while data-mining Eugene Go's answer to David's question in MSE, about Go's huge database to $a^4+b^4+c^4+d^4 = e^4$ with $e<10^7$. I was low-tech, using Google's free spreadsheet. Matej, was more high-tech and used an AI (Claude Fable). We got nowhere at first.\n\nAfter I edited Matej's answer, I suggested he ask Claude to find \"interesting\" linear relations between ALL terms $(a,b,c,d,e)$ similar to the Jacobi-Madden. Turns out Claude was up to the challenge. For example, given\n\n$$329580^4 + 32420^4 + 1911244^4 + 2018095^4 = 2339217^4$$\n\nClaude found the simple relation,\n\n$$\\frac{-a+e}{-b-c+d}= \\frac{-329580 + 2339217}{-32420 - 1911244 + 2018095}=\\color{red}{3^3}$$\n\n*How is a human being supposed to find that* in a sea of **45,000** solutions, not to mention the sign changes? And Claude found a total of 11 \"interesting\" relations, all of the form,\n\n$$\\frac{a+e}{b+c+d} = k^3$$\n\nwhere $k$ is a * small* rational number like $k=3,\\,5/4,\\,5/3,\\,5, 1/5,\\,20/29,$ etc. (\n\n**III. The equation**\n\nMatej reached out to me as he knew I did work on the $(4,1,3)$ and $(4,1,4)$. Their proof seems a bit complicated and technical, but it turns out there was a simple *algebraic identity behind it*. Sometimes, there may be a *quadratic form* or an *elliptic curve* involved, and it turns out the nine families were no different. Given,\n\n$$1+x^4+y^4+z^4 = \\big(1+k^3(x+y+z)\\big)^4$$\n\nDo the substitutions, $x = \\frac{p+q}2,\\; y=\\frac{p-q}2,\\; z = k+(h-1)(k+p)$, where $\\color{blue}h$ is a parameter to be given later. We get the biquadratic in $q$,\n\n$$q^4+Aq^2+B=0$$\n\nThe expressions for $(A,B)$ are quite long but Mathematica can easily find them. The solution $q$ is a bit simpler,\n\n$$q^2 = -3p^2\\color{red}{\\pm2}(k+p)\\sqrt{D^2}$$ $$D^2 = 8h (k^2 - k p + p^2) + 12 h^2(k^6 - p^2) + 8h^3 (k + p) (k^9 + p) + 2h^4( k^{12}-1) (k + p)^2$$\n\nand works for * any* three arbitrary parameters $(h,k,p)$.\n\nBut we wish to have rational $q$. Once $(h,k)$ are given, then the second equation fortunately is only a quadratic in $p$ to be made a square, easily done by *quadratic forms* using an initial solution. Substituting this quadratic form into the first equation which is also a quadratic in $p$, then one naturally gets $2\\times2 = 4$ or,\n\n$$q^2 = \\text{quartic polynomial}$$\n\njust a recipe for an *elliptic curve*.\n\n**IV. Example**\n\nThe smallest case Claude found was,\n\n$$13355^4 + 8010^4 + (-9498)^4 + 1530^4 = 14489^4$$\n\nFor simplicity, we divide everything by $13355^4$ and reduce the number of variables to just three,\n\n$$1+x^4+y^4+z^4 = \\big(1+27(x+y+z)\\big)^4$$\n\nwith $k^3 = 3^3 = 27$, so $k=3$ and $(x,y,z) = (1602/2671,\\, -9498/13355,\\, 306/2671)$. The *fixed* parameter $\\color{blue}h$ is defined as,\n\n$$h = \\frac{x+y+z}{k+x+y}$$\n\nHence, $h = -\\dfrac{1835}{1837} + 1 = \\dfrac{2}{1837}.\\,$ Since,\n\n$$(x,y) = \\left(1602/2671,\\; -9498/13355\\right) = \\left(\\frac{p+q}2,\\;\\frac{p-q}2\\right)$$\n\ntherefore $$(p,q) = \\left(-\\frac{1488}{13355},\\, \\frac{17508}{13355}\\right)\\quad$$\n\nand from initial $p$, we find polynomial $p$ is,\n\n$$p = -\\frac{1488}{13355} - \\frac{5511 (\\color{red}{48044056139 + 1242748 r})}{13355 (r^2 - 6190029774)}$$\n\nSubstituting polynomial $p$ into the formula in the previous section and using the *negative* case of $\\color{red}{\\pm2}$, we get a quartic in $r$ to be made the square $q^2$.\n\n**V. Summary**\n\nNow that we have $(p,q)$, we get the \"simple\" identity,\n\n$$1+x^4+y^4+z^4 = \\big(1+27(x+y+z)\\big)^4$$\n\nwhere $(x,y,z)$ are,\n\n$$(x,y) = \\left(\\frac{p+q}2,\\;\\frac{p-q}2\\right)$$\n\n$$z = 3+\\left(\\frac{2}{1837}-1\\right)(3+p)$$\n\nand $(p,q)$ are,\n\n$$p = -\\frac{1488}{13355} - \\frac{5511 (\\color{red}{48044056139 + 1242748 r})}{13355 (r^2 - 6190029774)}$$\n\n$$q = \\frac{3\\sqrt{D^2}}{13355\\, (r^2 - 6190029774)}$$\n\n$$D^2 = -(18958669594580211381729967107 + 1093490321304049798772416r + 15172317493269316128r^2 + 3306770731944r^3 + 35534992r^4)$$\n\nThis is valid for * any* $r$ except $r^2 - 6190029774=0.$ To find a rational point on $D^2=\\text{quartic}$, equate the red portion above to zero to get r =-48044056139/1242748 and we recover the original $(p,q)$,\n\n$$(p,q) = \\left(-\\frac{1488}{13355},\\, \\frac{17508}{13355}\\right)\\quad$$\n\nBut since $D^2=\\text{quartic}$ is birationally equivalent to an *elliptic curve with positive rank*, then there are infinitely many rational $r_n$ yielding *different* $(p,q)$, hence *different* $(x,y,z)$. For example, using the old tangent-chord method, we find $73$-digit solutions which surely can't be the next smallest.\n\n*Note*: This approach can be generalized to * all* nine families.\n\n**VI. Question**\n\n**Q:** The rational point $r_1$ seems to have a large height. Can we find smaller ones and find \"small\" primitive solutions to,\n\n$$a^4+b^4+c^4+d^4 = (a+27b+27c+27d)^4 = e^4$$\n\nwith $e<10^{20}$ using the elliptic curve above, or related ones?\n\n*Note*: For info, a table of 63 known solutions (mostly by Seiji Tomita) for the Jacobi-Madden $a^4+b^4+c^4+d^4 = (a+b+c+d)^4 = e^4$ with $e<10^{18}$ is in [MSE post](https://math.stackexchange.com/q/4864970/4781).", "url": "https://wpnews.pro/news/more-small-solutions-to-a-new-infinite-family-a-4-b-4-c-4-d-4-a-27b-27c-27d-4", "canonical_source": "https://mathoverflow.net/questions/514415/more-small-solutions-to-a-new-infinite-family-a4b4c4d4-a27b27c27d", "published_at": "2026-08-18 10:54:44+00:00", "updated_at": "2026-08-22 06:13:11.085354+00:00", "lang": "en", "topics": ["artificial-intelligence", "ai-research"], "entities": ["Matej Veselovac", "Tito Piezas III", "Anthropic", "Claude", "Eugene Go", "Jacobi-Madden"], "also_reported_by": [], "alternates": {"html": "https://wpnews.pro/news/more-small-solutions-to-a-new-infinite-family-a-4-b-4-c-4-d-4-a-27b-27c-27d-4", "markdown": "https://wpnews.pro/news/more-small-solutions-to-a-new-infinite-family-a-4-b-4-c-4-d-4-a-27b-27c-27d-4.md", "text": "https://wpnews.pro/news/more-small-solutions-to-a-new-infinite-family-a-4-b-4-c-4-d-4-a-27b-27c-27d-4.txt", "jsonld": "https://wpnews.pro/news/more-small-solutions-to-a-new-infinite-family-a-4-b-4-c-4-d-4-a-27b-27c-27d-4.jsonld"}}