Check Proof by Contradiction Please check the following proof.
Result to Prove: If $a$ and $b$ are odd integers, then $4\not| (a^2+b^2)$
Assume, to the contrary, that there exists two odd integers $a$ and $b$ such that $4|(a^2+b^2)$. 
Then, $(a^2+b^2)=4n$ for some $n\in\Bbb Z$. 
Let $a=2x+1$ and $b=2y+1$ for some $x,y\in \Bbb Z$.
$a^2+b^2=(2x+1)^2+(2y+1)^2=4x^2+4y^2+4x+4y+2=2(2x^2+2y^2+2x+2y+1)=4n$. 
$\implies 2x^2+2y^2+2x+2y+1=4n$
$\implies 2(x^2+y^2+x+y)+1=2n$.
Since $(x^2+y^2+x+y), n\in\Bbb Z$, $2(x^2+y^2+x+y)+1$ is odd and $2n$ is even, which is a contradiction.

$2(2x^2+2y^2+2x+2y+1)=4n$ does not imply $2x^2+2y^2+2x+2y+1=4n$, but $2x^2+2y^2+2x+2y+1=2n$. Other than that, it's fine.
