Can the new infinite family $a^4+b^4+c^4+d^4 = (a+27b+27c+27d)^4$ be split into two quadrics?

In 2008, Jacobi-Madden found (essentially by data-mining the 25 smallest solutions) that

$$a^4+b^4+c^4+d^4 = (a+b+c+d)^4 =e^4$$

was solvable and in fact a member of an infinite family.

In early August 2026, Matej Veselovac and I were data-mining the first 45,000 smallest solutions of Eugene Go's database. Since Matej had access to Anthropic's AI Claude Fable, I suggested he ask it to find "interesting" linear relations between the $(a,b,c,d,e)$ like if the coefficients are small.

It seems machines now know what is "interesting", since among the hundreds of possible linear relations of the 45,000 solutions, it found

$$329580^4 + 32420^4 + 1911244^4 + 2018095^4 = 2339217^4$$

and decided that,

$$\frac{-a+e}{-b-c+d}= \frac{-329580 + 2339217}{-32420 - 1911244 + 2018095}=27=\color{red}{3^3}$$

was "interesting". Thus,

$$a^4+b^4+c^4+d^4 = (a+27b+27c+27d)^4=e^4$$

was also solvable. Only 2 "small" solutions were found in the database but, by using an elliptic curve, it turns out they are members of an infinite family. More details in this MO question.

We wish to find an alternative method to find smaller solutions by considering quadrics (if it is possible) and comparing it to similar equations.

I. For $a^4+b^4+c^4=d^4$

Can also be expressed as,

$$(p + r)^4 + (p - r)^4 + s^4 = q^4$$

In 1998, Noam Elkies completely solved this as an intersection of two quadrics,

$$-(3 m^2 - 8m + 6) p^2 + 2 (m^2 - 2) p q - 2 m q^2 = (m^2 + 2) r^2\tag1$$

$$-4 (m^2 - 2) p^2 + 8 m p q + (m^2 - 2) q^2 = (m^2 + 2) s^2\tag2$$

for some parameter $m$. Eliminating $m$ using resultants, easily done now with Mathematica, one recovers the original equation with a trivial factor. Sample computation by WolframAlpha is here.

II. For $a^4+b^4+c^4+d^4 = (a+b+c+d)^4$

Can also be expressed as,

$$(p-2q + r)^4 + (p-2q - r)^4 + (q + s)^4 + (q - s)^4 = (2p - 2q)^4$$

Jacobi-Madden (JM) also solved this as an intersection of two quadrics,

$$(m^2-7) p^2 + 24pq-24q^2= (m^2+1) r^2\tag3$$

$$8mp^2-24mpq - 3(m^2 - 8m + 1) q^2 = (m^2 + 1) s^2\tag4$$

for some parameter $m$.

III. For $a^4+b^4+c^4+d^4 = \big(a+k^3(b+c+d)\big)^4$

Can also be expressed as,

$$(p - k^3q)^4 + (q - r)^4 + (q - s)^4 + (r + s)^4 = (p + k^3q)^4$$

Note: For simplicity, we can focus on $k^3=27$ as in the title.

IV. Question

Q: Can the third equation be also solved as an intersection of two quadrics for some parameter $m$ like the first two? (After all, the Jacobi-Madden is just $k=1$ while the new one is $k=3$.)

If yes, it may give an alternative elliptic curve with smaller coefficients, hence yield "small" solutions to the equation in the title.

If no, the answer would stil be interesting, as what makes this different from the first two?