Prove this proposition by cases When i try to prove this proposition by cases 
$$x^2 + w^2 + y^2 = z^2$$
Where $x$, $y$, $w$ are positive integers .. $z$ is even if and only if $x$, $y$ and $w$ are even
I represent even as $2i$
And odd as $2i + 1$
When i try the case that 2 of them are odd and the another is even 
I get $(4i^2 + 4i + 4i^2 + 4i + 2)$
And that is an even .. am i wrong ? Or the proposition is wrong ?
Thank you

The proposition is correct, and you are correct that $4i^2 + 4i + 4i^2 + 4i + 2 = 8i^2 + 8i + 2$ is an even number, but that sum is equal to $z^2$, not $z$.
There is no $z$ that squares to this quantity, even or odd -- something we can see by trying to factor it and failing. And we know such a $z$ isn't even, anyway, since (like Will Jagy's comment says) the square of an even number must be a multiple of $4$.
(Proof: $(2i)^2 = 4i^2$).
