Solutions to $a^4+b^4+c^4 = d^4+e^4$ with $d\neq e$?

(Updated with a computer search.)

I. Question

We seek to find infinitely many primitive solutions to,

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

where $d \color{red}{\ne} e$. The most well-known case when $d = e$ is,

$$a^4+b^4+(a+b)^4 = 2(a^2+ab+b^2)^2$$

where one then solves $a^2+ab+b^2 = z^k$ for $k=2$. (In fact, it can be solved for any positive integer $k$.) But there are infinitely many solutions where $d\pm e \neq 0$,

$$(6n)^4 + (4 - 3 n^4 - n^8)^4 + (4 n - n^9)^4 =\\ (4 + 3 n^4 - n^8)^4 + (2 n + n^9)^4$$

$$(1 - 2 n^4 - n^8)^4 + (2 n - 2 n^9)^4 + (3 n^3 - n^{11})^4 =\\ (1 + 2 n^4 - n^8)^4 + (n^3 + n^{11})^4$$

by Gerardin and Norrie, of deg-$9$ and deg-$11$, respectively. (Are there any of smaller degree?)

II. Numerical solutions

They seem quite rare compared to the case $d=e$. A quick Mathematica search yields only $14$ primitive solutions with all terms $x_i<100$, the smallest $10$ being,

3^4 + 20^4 + 26^4 = 7^4 + 28^4
3^4 + 26^4 + 35^4 = 17^4 + 37^4  
1^4 + 25^4 + 42^4 = 17^4 + 43^4
25^4 + 26^4 + 42^4 = 37^4 + 38^4
7^4 + 29^4 + 50^4 = 21^4 + 51^4
5^4 + 42^4 + 78^4 = 51^4 + 76^4
25^4 + 76^4 + 60^4 = 35^4 + 82^4  
17^4 + 52^4 + 80^4 = 48^4 + 81^4 
39^4 + 79^4 + 60^4 = 51^4 + 83^4  
4^4  + 57^4 + 85^4 = 11^4 + 89^4

III. Elliptic curve 1

The $3$rd has symmetries we can exploit, namely,

$$1^4 + (a + b)^4 + (2 b)^4 = (-a + b)^4 + (2 b + 1)^4$$

Expanding, this reduces to just a quadratic in $b$,

$$a^3 - 3 b + (a - 4) b^2 = 1$$

where its discriminant $D_1$ must be made a square $D_1 = 4 (a^3 - 1)(4 - a) + 9 = y^2$. From initial rational point $a=4$, we can find an infinite more rational $a$.

IV. Elliptic curve 2

The $4$th also has symmetries we can exploit,

$$(a - 2d)^4 + (a + 2d)^4 + (4c)^4 = (b - 2d)^4 + (b + 2d)^4$$

Expanding, this reduces to a quadratic in $d$,

$$a^4 - b^4 + 128 c^4 - 24 (-a^2 + b^2) d^2=0$$

Its discriminant $D_2$ must be made a square,

$$D_2 = 24 (-a^2 + b^2)(a^4 - b^4 + 128 c^4) = y^2$$

which is only a quartic in $c$. From the rational point $(a,b)=(17,\,25)$, we find initial rational point $c = 7$ then an infinite more rational (but not integral) $c$.

Note 1: A computer search for small integer $(a,b,c)$ with $\text{GCD}(a,b,c)=1$ yields,

(a,b,c) = (7, 31, 19)
(a,b,c) = (8, 16, 11)
(a,b,c) = (8, 20, 7)
(a,b,c) = (17, 25, 7)

though each of these integer $a<b<200$ will yield infinitely many rational $c$.

Note 2: To find useful symmetries on the other solutions was a bit more elusive.

V. Summary

So can we find more polynomial parameterizations or elliptic curves for $a^4+b^4+c^4 = d^4+e^4$ where $d\neq e$?

P.S. The third solution appears in Guruprasad's 2025 MO post. There are only three known with a unit term on the LHS,

$$1+25^4+42^4=17^4+43^4\\ 1+405^4+1786^4=253^4+1787^4\\ 1+3870^4+11954^4=6953^4+11632^4$$

the first two by Guruprasad, and the third by Peter Mueller. Mueller points out in his answer that if there is a fourth, then it must be enormous. There doesn't seem to be any known solution if the unit term is on the RHS,

$$a^4+b^4+c^4 = d^4+1$$