Finding all integral solutions of $x^4+y^4+z^4-w^4=1995$

Question

Find all the integral solutions of

$$x^4+y^4+z^4-w^4=1995.$$

Please elaborate the solution. I tried but can't understand what to do.

Answer 1

Fourth powers are either of $$ 0,1 \pmod {16}. $$ The possibilities are $$ x^4 + y^4 + z^4 - w^4 \equiv 15,0,1,2,3 \pmod {16}. $$ However, $$ 1995 \equiv 11 \pmod {16}. $$

Answer 2

I just figured out that 1995 is a multiple of 5 and by fermat's theorem N^(5-1)=1 mod 5 thus

LHS x^4+y^4+z^4-w^r=1+1+1-1=3 mod 5

However RHS is 0 mod 5 thus this is not possible

So now one condition is left when x, y, z, w are multiples of 5

Thus then LHS will have 5^4 as its factor which will not completely divide 1995 thus this we will have no solution in integer