Since AI (Claude) just helped find the new infinite family $a^4+b^4+c^4+d^4 = (a+27b+27c+27d)^4$, should we expect more similar discoveries soon?

Matej Veselovac, using the AI Claude, discovered a new infinite family of solutions to the equation a^4+b^4+c^4+d^4 = (a+27b+27c+27d)^4, along with eight additional families, by analyzing over 45,000 primitive solutions from a database with e<10^7. The discovery extends the 2008 Jacobi-Madden family and demonstrates AI's utility in data-mining large mathematical 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. 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 to find patterns, it seems AI can prove useful when data-mining an ocean 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 $$\frac{a+e}{b+c+d} = k^3$$ 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. The solution $q$ is the important point, $$4q^2 = -12p^2\color{red}{\pm8} k+p \sqrt{D^2}$$ $$D^2 = 8h k^2 - k p + p^2 + 12 h^2 k^6 - p^2 + 8h^3 k + p k^9 + p + 2h^4 k^{12}-1 k + p ^2$$ 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 parameter $h$ is defined as, $$h = \frac{x+y+z}{n+x+y}$$ Hence, $h = -1835/1837 + 1 = 2/1837.\,$ Since, $$ x,y = \left 1602/2671,\; -9498/13355\right = \left \frac{p+q}2,\;\frac{p-q}2\right $$ therefore $ p,q = -1488/13355, 17508/13355 $ and polynomial $p$ is, $$p = -\frac{1488}{13355} + \frac{5511 48044056139 + 1242748 r }{13355 r^2 - 6190029774 }$$ 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, $$q^2 = -R^2 18958669594580211381729967107 + 1093490321304049798772416r + 15172317493269316128r^2 + 3306770731944r^3 + 35534992r^4 $$ where $R = \dfrac{3}{13355 r^2 - 6190029774 }$. Let $r =-48044056139/1242748$ and we recover, $$q^2 = \left \frac{17508}{13355}\right ^2$$ This quartic is birationally equivalent to an elliptic curve with positive rank , hence there are infinitely many rational $r n$. This method can be generalized to all nine families. IV. Question Q: Can AI help find new infinite families $a^4+b^4+c^4+d^4 = e^4$ were the terms are now in a non-linear relation? In general, given a Diophantine equation with thousands of solutions, will we reach a point where AI can find patterns in the data that the human mind can't, and we just step in to formally confirm the pattern? P.S. If the question is too difficult, I'll be happy to be given the first few small 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. I don't think that $r n$ was the smallest one while the table of 63 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 .