{"slug": "more-solutions-to-a-new-infinite-family-a-4-b-4-c-4-d-4-a-27b-27c-27d-4", "title": "More solutions to a new infinite family $a^4+b^4+c^4+d^4 = (a+27b+27c+27d)^4$?", "summary": "Matej Veselovac, using the AI Claude to data-mine Eugene Go's database of about 45,000 primitive solutions to a^4+b^4+c^4+d^4 = e^4 with e<10^7, discovered a new infinite family of solutions to a^4+b^4+c^4+d^4 = (a+27b+27c+27d)^4, along with eight more families where k^3 ≠ 27. The discovery extends the Jacobi-Madden family from 2008 and demonstrates AI's utility in finding patterns in large datasets.", "body_md": "Back 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\nas well as **eight** more where $k^3 \\neq 27$. 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 (Eugene Go's database) to find patterns, it seems AI can prove useful when data-mining a sea of data.\n\n**I. The Story**\n\nMatej and I met while data-mining Eugene Go's answer to David's question in MSE, about a database of around 45,000 primitive solutions 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, on the other hand, was more high-tech and used an AI (Claude). We got nowhere.\n\nAfter I edited Matej's answer, I suggested he ask Claude to 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?* It is 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 * tiny* rational number like $k=3,\\,5/4,\\,5/3,\\,5,$ etc. (\n\n**II. 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*. In my posts, there's always a *quadratic form* or an *elliptic curve*, 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 quickly find them. It has solution $q$,\n\n$$4q^2 = -12p^2\\color{red}{\\pm8}(k+p)\\sqrt{D^2}$$ $$D^2 = 8h (k^2 - k p + p^2) + 12 h^2(k^6 - p^2) + 8h^3 (k + p) (k^9 + p) + 2h^4( k^{12}-1) (k + p)^2$$\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\nwhich is just a recipe for an *elliptic curve*.\n\n**III. Example**\n\nThe smallest case Claude found was,\n\n$$1+x^4+y^4+z^4 = \\big(1+27(x+y+z)\\big)^4$$\n\nso $k=3$ 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 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 $\\pm8$, we get the quartic in $r$ to be made a square,\n\n$$q^2 = -R^2\\times(18958669594580211381729967107 + 1093490321304049798772416r + 15172317493269316128r^2 + 3306770731944r^3 + 35534992r^4)$$\n\nwhere $R = \\dfrac{3}{13355\\, (r^2 - 6190029774)}.$\n\nTo find a rational point on $q^2=\\text{quartic}$, 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\nTo find the original $(x,y,z)$, recall that $(x,y) = \\left(\\frac{p+q}2,\\;\\frac{p-q}2\\right)$ and $z$ can be recovered from,\n\n$$1+x^4+y^4+z^4 = \\big(1+27(x+y+z)\\big)^4$$\n\nHowever, since $q^2=\\text{quartic}$ 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**IV. Question**\n\n**Q:** The rational point $r_1$ seems to have a large height. Can we find the first few 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 a related one?\n\n*Note*: Ironically, a smaller $r_n$ may yield a *slightly* larger $(x,y,z)$ than the \"seed\" solution. And for info, a table of 63 known solutions for the Jacobi-Madden $a^4+b^4+c^4+d^4 = (a+b+c+d)^4 = e^4$ with $e<10^{18}$ is [here](https://math.stackexchange.com/q/4864970/4781).)", "url": "https://wpnews.pro/news/more-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-solutions-to-a-new-infinite-family-a4b4c4d4-a27b27c27d4", "published_at": "2026-08-18 10:54:44+00:00", "updated_at": "2026-08-18 16:14:49.099921+00:00", "lang": "en", "topics": ["artificial-intelligence", "ai-research"], "entities": ["Matej Veselovac", "Claude", "Eugene Go", "Jacobi-Madden"], "alternates": {"html": "https://wpnews.pro/news/more-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-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-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-solutions-to-a-new-infinite-family-a-4-b-4-c-4-d-4-a-27b-27c-27d-4.jsonld"}}