Question on positive integer solutions of $x^4-y^4=z^2$

Question

Suppose the equation $x^4-y^4=z^2$ has solution(s) in positive integers. Then show that the least $x$ value of these solutions is odd.

Here is my attempt using contradiction
let $x=2k$ $$(2k)^4-y^4 = z^2$$

Any hint/help on how one should go about finding a contradiction ? Thanks!

score 5 · Accepted Answer · answered Apr 08 '15 at 11:47

5

For the least positive solution $(x,y)=1$

So both of $x,y$ can not be even

If $x$ is even and $y$ is odd

$x^4\equiv0\pmod4, y^4\equiv1\pmod4\implies x^4-y^4\equiv-1\pmod4$

But $z^2\equiv0,1\pmod4\not\equiv-1$

answered Apr 08 '15 at 11:47

lab bhattacharjee

that looks brilliant! also i think that mod 4 trick works for showing $x^2-y^2 = z^2 $ and $(x,y)=1$ implies $x$ is odd and $y$ is even. thank you so much – drae Apr 08 '15 at 11:51

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