The theorem for the question is as follows:
The positive integers $(a, b, c)$ are a primitive Pythagorean triple with an odd if and only if $a = m^2 - n^2$, $b = 2mn$, and $c = m^2 + n^2$ , where $m, n$ are relatively prime positive integers, not both odd, and $m > n$.
To show that a and b cannot be both odd or both even, do I need to show that either $m^2 - n^2$ or $2mn$ is odd and then show that the other must be even? How would I begin to show that one or the other is odd or even?
Side-A is always odd, if primitve, and can be any odd number greater than one so $m$ and $n$ must be of opposite parity or $m^2-n^2$ would be even.
Side-B is always even, because $B=2mn$, and is a multiple of four unless $(m,n)$ are both odd and therefore non-primitve such as $F(3,1)=(8,6,10)$.
Side-C must be odd because it's square is the sum of an odd square and an even square.
If $(m,n)$ have a common factor $(f)$, then $GCD(A,B,C)>1$. For example \begin{equation} (m,n)=(fq,fs)\\ \implies A=f^2q^2-f^2s^2=f^2(q^2-s^2)\\ \qquad B=2fqfs=f^2(2qs)\\ \qquad C=f^2q^2+f^2s^2=f^2(q^2+s^2) \end{equation} Above, we can see that if $(m,n)$ are not mutually prime, then $GCD(A,B,C)=f^2$.