All integer solutions​ of $x^4 + y^4 + z^4 - w^4= 1995$ This question is in the book 'The thrill and challenge of precollege mathematics'. I intend to attack this problem using Fermat's Little Theorem (FLT).
Notice that each term on LHS must be either of the form $5k$ or according to FLT $5k +1$ but the if $x^4$ is of the form $5k$ then $x=5c$ for some integer c. I can't think of any think after this.
I have also used the property that any fourth power when divided by $4$ leaves remainder $1$ if it is odd and $0$ if it is even. Using this I think there are only $2$ things possible. Either, $x,y,z$ are odd and $w$ is even or $x,y,z$ even and $w$ odd.
Thanks in advance. Don't give me answer tho just tell me which concept to apply, or if this is right approach, the how to attack the problem.
 A: You can use a stronger result about fourth powers: $$x^4 \equiv 0,1 \mod 16$$
A: We know that $7|1^4 +2^4 + 4^4$ and $1955= 3\times5\times7\times19$
Suppose $w>x>y>z$ and we have $x=w-a, y=w-b, z=w-c$; we may write:
$$(w-a)^4+(w-b)^4+(w-c)^4-w^4=1995$$
For a primitive solution let $z=7$ (a factor of $1995$) we must have:
$$7|a^4+b^4+c^4$$
Then comparing with first relation we get $a=1, b=2,c=4$ and we have:
$$6^4 +5^4 + 3^4 - 7^4 = 399$$
But: $1995 = 5\times 399$
Therefore equation $x^4 +y^4 +z^4 - w^4 = 1995$ has no integer solution.
Generally we have to look for one or more common divisors between rhs and lhs of equation by checking a factor like $\alpha$  of rhs in equation, Suppose $w$ is a primitive solution and we have following conditions:
$(w± a)^4+(w± b)^4+(w± c)^4-w^4=\alpha.\beta.\gamma. . .$
$w|a^4+b^4+c^4$ and $[w,(a^4+b^4+c^4)]=\alpha$
We have integer solutions if these conditions gives all factors in rhs and we have no integer solutions if it does not.In our case 7 does not give the factor 5 in rhs, therefore the equation has no integer solution.
