<s>[INST] Do we simplify the Proof by Contradiction? Prove the following by contradiction:

Suppose $a,b\in\mathbb{Z}$. If $4|\left(a^2+b^2\right)$, then $a$ and $b$ are not both odd (in other words, $a$ and $b$ are even)
So, I did this:

Assume $a$ and $b$ are odd
  
  Let $a={2k+1}$
  
  Let $b={2l+1}$
  

  $4|\left((2k+1\right)^2+\left(2l+1\right))^2$
  
  $\left(2k+1\right)*\left(2k+1\right)=4k^2+4k+1$
  
  $\left(2l+1\right)*\left(2l+1\right)=4l^2+4l+1$
  

  $4|\left(\left(4k^2+4k+1\right)+\left(4l^2+4l+1\right)\right)$
  

  For this, do we still, simplify it
  
  Or, what do you do after that??

 [/INST] Because 
$$
a^2+b^2=(a-b)^2+2ab,
$$
if $a^2+b^2$ is even, then $a$ and $b$ must share parity. But if $a$ and $b$ are both odd, then $(a-b)^2$ is divisible by $4$ whereas $2ab$ is not, so it must be that $a$ and $b$ are both even.
</s>