Proof by contradiction proving both numbers are not odd. I have to do a proof by contradiction:

Suppose $a,b,\in\mathbb{Z}$. If $4| (a^2 + b^2)$ then a and b are not both odd.

So far I know that I need to prove that if $4|(a^2+b^2)$ then a and b are both odd.

I would use the definition of an odd number $(2k+1), k\in\mathbb{Z}$

I'm a little but unsure of where to go from here. Any help would be appreciated.
If both are odd then $a=2n+1$, $b=2m+1$ and $$a^2+b^2=4(n^2+n+m^2+m)+2$$ which is never divisible by $4$.
Let $a = (2k+1)$ and let $b=2k'+1$ for some k in the naturals. If you assume both are odd, and you don't get a factor of 4, then clearly 4 doesn't divide $(a^2 + b^2)$ when both $a,b$ are odd.