$z^2=x^2+y^2$ Prove that $4\mid xyz$ ($xyz$ is divided by $4$) $z^2=x^2+y^2$ where $x,\ y,\ z$ - integers
Prove that $4\mid xyz$ ($xyz$ is divided by $4$)
All possible rest in divided by $4$ in this case is $1$. That's all I noticed.

If an integer $n$ is not divisible by $4$, then it is equal to $1,2,3,5,6,$ or $7$ mod $8$. So $n^2$ is equal to $1$ or $4$ mod $8$.
Hence if neither $x$ nor $y$ is divisible by $4$, then $z^2 = x^2+y^2$ is equal to $0, 2,$ or $5$ mod $8$; in particular, $z^2$ is not equal to $1$ or $4$ mod $8$. And so $z$ must be divisible by $4$.
