If $a^2+b^2=c^2$ then $a$ or $b$ is even. I am having trouble proving this directly. I know that it is easy to prove by contradiction by assuming both $a$ and $b$ to be odd, but how should I start to try to prove this? This is a homework problem and I am just looking for help getting started, not the full answer.

Unrolling @ajotatxe's one-sentence proof
Auxiliary claim: Every odd perfect square is of the form $4m + 1$.
Proof. $(2k + 1)^2 = 4(k^2 + k) + 1 = 1 (\text{ mod } 4)$. Also note the elementary fact that an integer is odd iff its square is odd. $\quad \quad \quad \Box$
Now, if $c^2 = a^2 + b^2$ with both  $a$ an $b$ odd, then $c^2 = 1 +  1 = 2 (\text{ mod } 4)$, a contradiction, since the only squares modulo $4$ are $0$ and $1$. $\quad \quad \quad \Box$
