For a primitive pythagorean triple $(a, b, c)$, prove $(c+b), (c-b)$ are both squares. Problem
For a primitive pythagorean triple $(a, b, c)$ (with $a$ odd, $b$ even), prove $(c+b), (c-b)$ are both squares.
My thoughts
Having $a$ be odd, gives $a^2$ as odd also.
From $a^2 + b^2 = c^2$ we have $a^2 = (c-b)(c+b)$.
I have figured out that $\gcd(c-b, c+b) = 1$, which seems relevant.
From the fundamental theorem of arithmetic, we now know that the prime factorizations of $c-b, c+b$ will have no common factors. This also seems relevant.
But I'm having a hard time tying all this together.
Any help would be appreciated!

Given $A=m^2-n^2\quad B=2mn\quad C=m^2+n^2$
$$C-B=m^2+n^2-2mn=(m-n)^2$$
$$C+B=m^2+n^2+2mn=(m+n)^2$$
