If $2xy$ is a perfect square, then $x^2+y^2$ cannot be I ask a question that is probably never proved a conjecture: if $x$ and $y$ are two natural numbers $> 1$ such that $2xy = N^2$ ( double their product is a perfect square ), x and y cannot be part of a Pythagorean triple , that is, there is no natural $M$, $M > 1$ such that $x^2 + y^2 = M^2$.

This conjecture is true.
Assume for contradiction such $x$, $y$ exist. We may assume that $\gcd(x,y) = 1$ without loss of generality. Then according to $x^2+y^2=M^2$ there exist $m$ and $n$ of different parity and $\gcd(m,n) = 1$ such that $x = m^2-n^2$ and $y = 2mn$ (Euclid's Formula), so we would need $mn(m-n)(m+n)$ to be a square.
As $m$, $n$, $m-n$, $m+n$ are all pairwise coprime (note exactly one of them is odd), this can only happen if each is a perfect square; if $m = a^2$, $n = b^2$ then we get $c^2 = a^4-b^4$, but this is not possible.
