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!
Lord Shark's answer in the comments is probably the most elementary one. An alternative answer is to use the classification of primitive Pythagorean triples.
There are coprime, natural numbers $m,n$ (one of them even, the other odd) such that $$a=m^2-n^2\\b=2mn\\c=m^2+n^2$$ It's a bit of work to prove this, but once we have this result, the answer is immediate: $$c-b=(m-n)^2\\c+b=(m+n)^2$$