Prove that either $a$ is even or $b$ is even? Prove that if $a$, $b$, and $c$ are all positive integers such that $a^2$ + $b^2$ = $c^2$ then $a$ or $b$ must be even.
One way I was taught to handle implications with an or statement was to prove that if the first part is false, then the second part must be true. What I mean by this is that I can prove this implication to be correct if I can show that if $a$ is odd, then $b$ must be even. My problem is that I don't have much to work with. If $a$ is odd, then I could split the proof into 2 cases, one where $c$ is odd and the other where $c$ is even. I couldn't seem to get anywhere with that though.
 A: Suppose they are both odd. Then we get, for some integral $m,n$, $4m^2 + 4m + 4n^2 + 4n + 2 = c^2$. $c$ must thus be even, and hence it can be expressed as $c=2b$ for integral $b$. So we get 
$4m^2 + 4m + 4n^2 + 4n + 2 = 4b^2$ for some $b$. Dividing both sides by four, we see $b^2$ is not integral, which implies $b$ is not integral. This is a contradiction. 
A: If $a=2n+1$ and $b=2m+1$ are odd, then 
$$
a^2 = 4n^2+4n+1 = 4n(n+1)+1 \equiv 1 \mod 8
$$
because either $n$ or $n+1$ must be even. Similarly, $b^2\equiv 1\mod 8$. Now
\begin{align*}
a^2 + b^2 = 4n^2+4n+1 + 4m^2+4m+1 \equiv 2 \mod 8.
\end{align*}
But $2$ is not a square modulo $8$. Hence, there exists no $c$ such that $a^2+b^2=c^2$.
A: The general parametrized solution to this diophantine equation is: $a = 2pq, b = p^2-q^2, c = p^2+q^2$. This shows that $a$ is even.
A: We can understand that what we want to prove is that 
$ (odd)^2 + (odd)^2 $ will not equal to $(even)^2$
(Sum of odd numbers will only be even)
Perfect odd squares are in the form 
$4n + 1$
Therefore the above expression (sum of two odd  square numbers) comes down to 
$ 4n + 4m + 2$
It can be proven that this will never be an even pefect square which is always a multiple of 4.
If it is, then
$(4n + 4m + 2)/4 $ 
will be an integer, but it is not, as it becomes
$ m + n + 0.5$
