<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] $$
(2k + 1)^2 + (2l + 1)^2 = 4\left(k^2 + l^2 + k + l\right) + 2
$$
This means that $\left(a^2 + b^2\right) = 4 \lambda + 2$, meaning that $4$ could not possibly divide $\left(a^2 + b^2\right)$, since there's always a remainder of $2$.
...by the way, this means that for $4$ to divide $a^2 + b^2$, both $a$ and $b$ must be even, since certainly if one is odd and the other even, you will get an odd.  Further, if they are both even, then $4$ definitely divides them.
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