Find all integral solutions

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.

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And you can not understand. Such equations can not be solved. Even for such simple equations really do not know how to write the formula. math.stackexchange.com/questions/494534/… And when you write the formula for some reason it erased. – individ May 26 '14 at 5:10
This damn bad Google translate. – individ May 26 '14 at 5:11

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, $$1995 \equiv 11 \pmod {16}.$$

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I don't understand your solution. Could you please elaborate on what you're doing here? – Nick May 26 '14 at 5:49
His solution is basically showing there are no integral solution sets to the original equation. He's working modulo 16. The first assertion is easy to show when you consider that integers are either odd $(2k+1)$ or even $(2k)$, and take their fourth powers mod 16. The only possible values are 1 or 0. The next step is to determine what possible values the entire expression can take mod 16. $x^4+y^4+z^4$ can be taken together giving 0,1,2 or 3. Subtracting off 0 & 1 for $w^4$ gives -1,0,1,2,3 as possibilities. $-1 \equiv 15 \pmod{16}$. The final step is to note that none can equal $11\pmod{16}$. – Deepak May 26 '14 at 6:11

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

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I think your proof has a hole. The LHS can be reduced to $3\pmod{5}$ only if 5 does not divide any of x, y or z. You've covered the all divisible by 5 possibility. But you haven't covered the case where some of them (but not all) are divisible by 5. Example: if $x=1, y = z= 0$ and $w = 1$ (mod 5), you'll have an issue because the LHS will be zero (mod 5). – Deepak May 26 '14 at 6:24
Thanks for pointing out the mistake that's what I like about stack exchange – user3650050 May 26 '14 at 6:25