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.

Note that fourth powers are always congruent to $0$ or $1$ mod $16$. This can be proved using case analysis. I'll list these cases:

• If $n$ is even, then $2 \mid n$ hence $16 \mid n^4$.
• If $n\equiv \pm1 \mod 16$, then $n^4 \equiv 1 \mod 16$
• If $n\equiv \pm3 \mod 16$, then $n^4 \equiv 3^4 \equiv 1 \mod 16$
• If $n\equiv \pm5 \mod 16$, then $n^4 \equiv 5^4 \equiv 1 \mod 16$
• If $n\equiv \pm7 \mod 16$, then $n^4 \equiv 7^4 \equiv 1 \mod 16$

Now, the left hand side can be $-1, 0, 1, 2$ or $3$ mod $16$, but $1995$ is congruent to $11$ mod $16$.

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 $(2k+1)^4\equiv 1\pmod{16}$.