Find all integral solutions of $ x^4 + y^4 + z^4 -w^4 = 1995 $ Find all integral solutions of $ x^4 + y^4 + z^4 -w^4 = 1995 $.  
Attempt:
From FLT it can be concluded that either all of $ x , y , z$ and $w$  are multiples of 5 ( which is not possible since that would lead to $ x^4 + y^4 + z^4 -w^4 $ being a multiple of $5^4$ which it is not since it's equal to 1995) or
$ w^4 $ and  exactly one of $x^4, y^4, z^4 $ are of the form $5k+1$ .
So, there can be 3 cases with each having either $x^4, y^4  $ or $ z^4 $, along with $w^4$ , of the form $5k+1$ .
I am unable to find the solutions from there.

Just an addendum to wythagoras fine (+1) answer. $x^4\in\{0,1\}\pmod{16}$ can be proved by observing that if $x$ is even, $x^4$ is clearly a multiple of $16$, while
$$ (2k+1)^4-1 = 2k(2k+2)(4k^2+4k+2) = 16\binom{k+1}{2}(2k^2+2k+1) $$
gives $(2k+1)^4\equiv 1\pmod{16}$.
