More solutions to a new infinite family $a^4+b^4+c^4+d^4 = (a+27b+27c+27d)^4$?

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.

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, $$a^4+b^4+c^4+d^4 = a+27b+27c+27d ^4$$ as 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. I. The Story Matej 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. After 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 $$329580^4 + 32420^4 + 1911244^4 + 2018095^4 = 2339217^4$$ 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? 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, $$\frac{a+e}{b+c+d} = k^3$$ where $k$ is a tiny rational number like $k=3,\,5/4,\,5/3,\,5,$ etc. II. 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 . In my posts, there's always a quadratic form or an elliptic curve , 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 quickly find them. It has solution $q$, $$4q^2 = -12p^2\color{red}{\pm8} k+p \sqrt{D^2}$$ $$D^2 = 8h k^2 - k p + p^2 + 12 h^2 k^6 - p^2 + 8h^3 k + p k^9 + p + 2h^4 k^{12}-1 k + p ^2$$ 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, $$q^2 = \text{quartic polynomial}$$ which is just a recipe for an elliptic curve . III. Example The smallest case Claude found was, $$1+x^4+y^4+z^4 = \big 1+27 x+y+z \big ^4$$ 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 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 $\pm8$, we get the quartic in $r$ to be made a square, $$q^2 = -R^2\times 18958669594580211381729967107 + 1093490321304049798772416r + 15172317493269316128r^2 + 3306770731944r^3 + 35534992r^4 $$ where $R = \dfrac{3}{13355\, r^2 - 6190029774 }.$ To 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 $, $$ p,q = \left -\frac{1488}{13355},\, \frac{17508}{13355}\right \quad$$ To 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, $$1+x^4+y^4+z^4 = \big 1+27 x+y+z \big ^4$$ However, 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. Note : This approach can be generalized to all nine families. IV. Question Q: The rational point $r 1$ seems to have a large height. Can we find the first few 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 a related one? 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 .