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 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.)