contradiction proof on divides Suppose a,b ∈ Z. If 4 | $(a^2 + b^2)$ then a and b are not both odd.
So, assuming that 4 | $(a^2 + b^2)$  and $a$ and $b$ are odd
this gives $4k=(2l+1)^2+(2u+1)^2$ for some $k,l,u\in z$
eventually leading to $4k=4(l^2+l+u)+2(u^2+1)$
The RHS is not a multiple of 4 when $u=2$ contradiction.
Is this valid, thanks.

It's not yet valid, because you haven't shown why the RHS cannot be a multiple of $4$. You cannot simply set $u=2$, because the $u$ you have is already determined by $b$, since $b=2u+1$.

To correct your proof, re-think how you got from
$$4k=(2l+1)^2 + (2u+1)^2$$
to 
$$4k = 4(l^2+l+u) + 2(u^2+1)$$
because I think you were a bit sloppy here.
