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 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 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.