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.

This is invalid. You have done a mistake in calculation.
RHS will be $4(l^2+u^2+l+u)+2$, which is not divisible by 4.
Hope it helps:)
