<s>[INST] Proof involving integer squares and parity Original proposition:

$a$, $b$ and $c$ are integers. Show that if $a^2+b^2=c^2$, at least
  one of $a$ and $b$ is an even number.

My attempt:
Attemplting proof by contradiction:
Show that there exist a contradiction between the claim that $a^2+b^2=c^2$
...and $a=2k+1, b=2k+1$ where $a\in\mathbb{Z}$ and $a\in\mathbb{Z}$:
Or show:
$$(2k+1)^2+(2l+1)^2\neq c^2$$
I converted the above to:
$$4(k^2+k+l^2+l)+2\neq c^2$$
Perhaps it is possible to show that $4(k^2+k+l^2+l)$ is a perfect square and hence $4(k^2+k+l^2+l)+2$ cannot be a perfect square?
 [/INST] Assume by way of contradiction $a$ and $b$ are odd, as you say we get that $$4(k^2+k+l^2+l)+2=c^2.$$ We have that $c^2$ is even thus $c$ is even, so $c=2m$ for some $m$. We get $$4(k^2+k+l^2+l)+2=4m^2.$$
divide both sides by two: 
$$2(k^2+k+l^2+l)+1=2m^2$$
This  is a contradiction since LHS is odd and RHS is even. 
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