# $x^4 + y^4 = z^2$

$x, y, z \in \mathbb{N}$,

$\gcd(x, y) = 1$

prove that $x^4 + y^4 = z^2$ has no solutions.

It is true even without $\gcd(x, y) = 1$, but it is easy to see that $\gcd(x, y)$ must be $1$

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Suppose that $x=ku$ and $y=kv$ for some integer $k$. Then $x^4+y^4=k^4(u^4+v^4)$, which is a square if and only if $u^4+v^4$ is a square. –  Brian M. Scott Jan 12 '13 at 20:03
@Brian: That looks like an answer to me... –  TMM Jan 12 '13 at 20:04
I said the part I cant solve is when GCD = 1. –  Bojan Serafimov Jan 12 '13 at 20:05
There is a proof here. –  P.. Jan 12 '13 at 20:25
post an answer. –  Bojan Serafimov Jan 12 '13 at 20:36

Suppose that $x^4+y^4=z^2$, where $z$ is the smallest positive integer for which there is a solution in positive integers. Then $(x^2,y^2,z)$ is a primitive Pythagorean triple, so there are relatively prime integers $m,n$ with $m>n$ such that $x^2=m^2-n^2,y^2=2mn$, and $z=m^2+n^2$.
Since $2mn=y^2$, one of $m$ and $n$ is an odd square, and the other is twice a square. In particular, one is odd, and one is even. Now $x^2+n^2=m^2$, and $\gcd(x,n)=1$ (since $m$ and $n$ are relatively prime), so $(x,n,m)$ is a primitive Pythagorean triple, and it must be $n$ that is even: there must be integers $a$ and $b$ such that $a>b$, $a$ and $b$ are relatively prime, $x=a^2-b^2$, $n=2ab$, and $m=a^2+b^2$. It must be $m$ that is the odd square, so there are integers $r$ and $s$ such that $m=r^2$ and $n=2s^2$.
Now $2s^2=n=2ab$, so $s^2=ab$, and we must have $a=c^2$ and $b=d^2$ for some integers $c$ and $d$, since $\gcd(a,b)=1$. The equation $m=a^2+b^2$ can then be written $r^2=c^4+d^4$.