Do we simplify the Proof by Contradiction?

Question

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??

Answer 1

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

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.

Answer 2

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.