Is this contradiction proof correct? Statement : suppose $a,b$ belongs to $\mathbb{Z}$ (integers). If $4/ (a^2 + b^2)$ then $a$ and $b$ are not both odd. Proof by contradiction: Assume that if $4/(a^2 + b^2)$ then a and b are both odd. If $a^2$ and $b^2$ is odd then by definition a and b must be odd too. It follows that $a^2$ (or $b^2$) $= (4k+1)^2$ ----- (unsure about this step) Then a (or b) $= 4k+1$ ---- (unsure about this step) So if $a$ and $b$ are both odd then this is a contradiction hence the supposition is false and the statement is true.

The proof by contradiction should start with

> assume that $a$ and $b$ are both odd and $4\mid (a^2+b^2)$

You can continue by saying

> write $a=2h+1$ and $b=2k+1$

and

> so that $a^2+b^2=4h^2+4h+1+4k^2+4k+1=\dots$

Now you can add

> and we have a contradiction, because this would imply $\dots$