x4 is congruent to either 0 or 1 mod 5.............Now, the RHS is congruent to 0 mod 5, so x4+y4+z4 = w4 (mod5)............And so, this is only possible when x,y,z,w are all multiples of 5............but that is not possible or else 1995 would be divisible by 625.............Hence, there are no integral solutions

@Henry U
–
Yeah you are right.............But I figured it out.......mod 5 is not a good option to look...........instead, try mod 16............that works fine.......:)

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## Comments

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TopNewestThere are no integral solutions.............Hint :- Look mod 16.............

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How!! Didn't understand?Please elaborate it!!

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x4 is congruent to either 0 or 1 mod 5.............Now, the RHS is congruent to 0 mod 5, so x4+y4+z4 = w4 (mod5)............And so, this is only possible when x,y,z,w are all multiples of 5............but that is not possible or else 1995 would be divisible by 625.............Hence, there are no integral solutions

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$x^4+y^4+z^4 \equiv w^4 \pmod 5$, there's also the possibility $1+0+0 \equiv 1 \pmod 5$.

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Any number to the power of 4 has a remainder of 0 or 1 when divided by 16.

So $x^4 + y^4 +z^4 - w^4$ can only have a remainder of 0,1,2, or 3 when divided by sixteen.

When 1995 is divided by 16 it leaves a remainder of 11.

Therefore there are no integral solutions.

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