<s>[INST] Number Theory equality 
Suppose that the equation $x^4+y^4=z^2$ has solutions $(x,y,z)\in\mathbb{Z}^{3+}$, let $c$ be the smaller $z$ for which $(x,y,c)$ is a solution. (1) If $x$ is even, prove $x^2=m^2-n^2, y^2=2mn, c=m^2+n^2$, $m,n$ relatively prime. (2) Prove $x=r^2 - s^2, n=2rs, m=r^2+s^2$, $r,s$ relatively prime.

(1) I have the following theorem: if $x^2+y^2=z^2$ then $x^2=2ab, y^2=b^2-a^2, c=b^2+a^2$.
Taking the $x^2$ to be $y$ in the latter, $y^2$ to be $x$ and $c$ to be $z$, then there is $m,n$ such that $x^2=m^2-n^2, y^2=2mn, c=m^2+n^2$.
(2) $x^2+y^2=m^2-n^2+2mn=(m-n)^2$. By means of the same theorem, there is $r,s$ such that $x=r^2-s^2, y=2rs, m+n=r^2+s^2$. This gives the equation for $x$, but the other two don't seem two follow, because it needs too be $n=y$ and if $m+n=r^2+s^2$ then $m=(r-s)^2$ instead of $m=r^2+s^2$.
 [/INST] you have $x^2+n^2=m^2$, so by the theorem, there are integers $r,s$ such that $x=r^2-s^2, n=2rs, m=r^2+s^2$.
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