(Note: See also the $a^4+b^4+c^4 = 1$ version in this old MSE post.)
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 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:
- 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?