# More small solutions to a new infinite family $a^4+b^4+c^4+d^4 = (a+27b+27c+27d)^4$? > Source: > Published: 2026-08-18 10:54:44+00:00 **Q:** We are looking for rational $(D,r)$, $$D^2=-(18958669594580211381729967107 + 1093490321304049798772416r + 15172317493269316128r^2 + 3306770731944r^3 + 35534992r^4)$$ with 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. **I. The first family** Back in the pre-AI years of 2008, Jacobi-Madden found the infinite family, $$a^4+b^4+c^4+d^4 = (a+b+c+d)^4$$ Just a few days ago, **Matej Veselovac** in an [MSE post](https://math.stackexchange.com/a/5146910/4781) found the **new** family, $$a^4+b^4+c^4+d^4 = (a+27b+27c+27d)^4$$ where $k^3$ 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. **II. The Story** Matej 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. After 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 $$329580^4 + 32420^4 + 1911244^4 + 2018095^4 = 2339217^4$$ Claude found the simple relation, $$\frac{-a+e}{-b-c+d}= \frac{-329580 + 2339217}{-32420 - 1911244 + 2018095}=\color{red}{3^3}$$ *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, $$\frac{a+e}{b+c+d} = k^3$$ where $k$ is a * small* rational number like $k=3,\,5/4,\,5/3,\,5, 1/5,\,20/29,$ etc. ( **III. The equation** Matej 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, $$1+x^4+y^4+z^4 = \big(1+k^3(x+y+z)\big)^4$$ Do 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 $q$, $$q^4+Aq^2+B=0$$ The expressions for $(A,B)$ are quite long but Mathematica can easily find them. The solution $q$ is a bit simpler, $$q^2 = -3p^2\color{red}{\pm2}(k+p)\sqrt{D^2}$$ $$D^2 = 8h (k^2 - k p + p^2) + 12 h^2(k^6 - p^2) + 8h^3 (k + p) (k^9 + p) + 2h^4( k^{12}-1) (k + p)^2$$ and works for * any* three arbitrary parameters $(h,k,p)$. But 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, $$q^2 = \text{quartic polynomial}$$ just a recipe for an *elliptic curve*. **IV. Example** The smallest case Claude found was, $$13355^4 + 8010^4 + (-9498)^4 + 1530^4 = 14489^4$$ For simplicity, we divide everything by $13355^4$ and reduce the number of variables to just three, $$1+x^4+y^4+z^4 = \big(1+27(x+y+z)\big)^4$$ with $k^3 = 3^3 = 27$, so $k=3$ and $(x,y,z) = (1602/2671,\, -9498/13355,\, 306/2671)$. The *fixed* parameter $\color{blue}h$ is defined as, $$h = \frac{x+y+z}{k+x+y}$$ Hence, $h = -\dfrac{1835}{1837} + 1 = \dfrac{2}{1837}.\,$ Since, $$(x,y) = \left(1602/2671,\; -9498/13355\right) = \left(\frac{p+q}2,\;\frac{p-q}2\right)$$ therefore $$(p,q) = \left(-\frac{1488}{13355},\, \frac{17508}{13355}\right)\quad$$ and from initial $p$, we find polynomial $p$ is, $$p = -\frac{1488}{13355} - \frac{5511 (\color{red}{48044056139 + 1242748 r})}{13355 (r^2 - 6190029774)}$$ Substituting 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 $q^2$. **V. Summary** Now that we have $(p,q)$, we get the "simple" identity, $$1+x^4+y^4+z^4 = \big(1+27(x+y+z)\big)^4$$ where $(x,y,z)$ are, $$(x,y) = \left(\frac{p+q}2,\;\frac{p-q}2\right)$$ $$z = 3+\left(\frac{2}{1837}-1\right)(3+p)$$ and $(p,q)$ are, $$p = -\frac{1488}{13355} - \frac{5511 (\color{red}{48044056139 + 1242748 r})}{13355 (r^2 - 6190029774)}$$ $$q = \frac{3\sqrt{D^2}}{13355\, (r^2 - 6190029774)}$$ $$D^2 = -(18958669594580211381729967107 + 1093490321304049798772416r + 15172317493269316128r^2 + 3306770731944r^3 + 35534992r^4)$$ This 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 $(p,q)$, $$(p,q) = \left(-\frac{1488}{13355},\, \frac{17508}{13355}\right)\quad$$ But 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. *Note*: This approach can be generalized to * all* nine families. **VI. Question** **Q:** The rational point $r_1$ seems to have a large height. Can we find smaller ones and find "small" primitive solutions to, $$a^4+b^4+c^4+d^4 = (a+27b+27c+27d)^4 = e^4$$ with $e<10^{20}$ using the elliptic curve above, or related ones? *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).