(***Note:*** See also the $a^4+b^4+c^4 = 1$ version in this old [MSE post](https://math.stackexchange.com/questions/509526/).)


The equation discussed in a paper by Jacobi and Madden,


$$a^4+b^4+c^4+d^4 = (a+b+c+d)^4 = z^4\tag1$$


or equivalently,


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


After some algebra, $(2)$ can be solved as an intersection of two rather simple quadric surfaces,


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


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


for some constant $m$. Given a known solution to $(1)$, $m$ can be recovered as,


$$m =\frac{3q^2+s^2}{p^2-r^2}\tag5$$


There are only two (?) known [primitive solutions](http://www.math.uni.wroc.pl/~jwr/eslp/414.txt) to $(1)$ with $z<220000$.


**I. Solution 1:**


$$(-2634)^4+5400^4+1770^4+955^4 = (-2634+5400+1770+955)^4=5491^4$$


From these $a,b,c,d$, after permutation one can get six distinct values for $m$,


$$m\_k =\frac{511}{450}, \, \frac{31^2}{61}, \, \frac{1423}{1098}, \, \frac{2521}{325}, \, \frac{1651}{126},\,\frac{1777}{1525}\tag6$$


which are, in fact, (courtesy of a comment by Jeremy Rouse),


$$m\_k =\frac{n\_1}{d\_1},\,\frac{n\_1+d\_1}{n\_1-d\_1},\,\frac{n\_3}{d\_3},\,\frac{n\_3+d\_3}{n\_3-d\_3},\,\frac{n\_5}{d\_5},\,\frac{n\_5+d\_5}{n\_5-d\_5}$$


For example, using $m\_2 =\frac{31^2}{61}$, and applying it to $(3),(4)$, we get,


$$448737 p^2 + 44652 p q - 44652 q^2 = 463621 r^2$$


$$234484 p^2 - 703452 p q - 687411q^2 = 463621 s^2$$ 


which has initial rational $p,q = -1906,\,\frac{1679}{2}$. Using an elliptic curve, we can then get an infinite more. 


**II. Solution 2:**


$$(-31764)^4+ 27385^4+ 48150^4+ 7590^4 = (-31764+27385+ 48150+7590)^4 = 51361^4$$


From these, we get,


$$m\_k =\frac{193}{18},\, \frac{211}{175},\, \frac{619}{450},\,\frac{1069}{169},\, \frac{1141}{666},\, \frac{1807}{475}\tag7$$


For example, using $m\_1 = \frac{193}{18}$, with initial $p,\,q = \frac{27187}{2}, -12087$.


**Question:**


1. Without knowing solutions 1 and 2 in advance, is there a systematic way to search for appropriate $m$ such that $(3),\,(4)$ has rational solutions? What other $m$ is there of small height not in the list of twelve above?



More elliptic curves for $a^4+b^4+c^4+d^4 = (a+b+c+d)^4$?
Given a rational number $m$, let $C\_{m}$ be the intersection of two quadrics given by (3) and (4) in the original question. The way to search for points is to test the curve $C\_{m}$ to see if it has local points. I did this for all $m$ with height $\leq 193$ (since I knew I'd find points on $C\_{193/18}$). There are only $18$ rational numbers for which $C\_{m}$ has points in $\mathbb{R}$, and $\mathbb{Q}\_{p}$ for $p = 2$ an all primes of bad reduction of the Jacobian of $C\_{m}$. (I did this computation in Magma, and it was complicated by the fact that Magma currently has a bug in its IsLocallySolvable command.)


(The $18$ rationals of height less than or equal to $193$ for which $C\_{m}$ has local points are $157/150$, $163/150$, $151/126$, $181/150$, $121/96$, $103/78$, $97/72$, $79/54$, $37/25$, $67/42$, $49/24$, $73/25$, $109/25$, $31/6$, $133/25$, $169/25$, $181/25$, and $193/18$.)


Searching for points on these $C\_{m}$ of height up to $10^{8}$ reveals points for $m = 157/150$, $m = 37/25$, $m = 31/6$ and the point on $m = 193/18$ that we already knew about. New solutions to (1) that arise from these points are
\begin{align\*}
 a &= 841263, b = -792940, c = 3852350, d = -44410\\
 a &= 35847220, b = -34122866, c = 53902630, d = 2542025\\
 a &= 39913670, b = -23859495, c = 15187700, d = 10116014,\\
\end{align\*}
which come from $m = 157/150$, $m = 37/25$ and $m = 31/6$, respectively. (I actually found two points on the $m = 31/6$ curve. The other point gives a solution to (1) which is a permutation of the $a = 35847220$ solution.)


It may be possible to prove that $C\_{m}$ has no points on it for other choices of $m$ using the fact that $C\_{m}$ is a four-cover of its Jacobian and using other techniques (the Cassels-Tate pairing, 3-descents, $L$-functions) to show that $C\_{m}$ represents an element of Sha. 


