An interesting question about greatest common divisor (gcd) of three positive integers which form a primitive Pythagorean triple

Question

Suppose $a$, $b$, and $c$ are three distinct integers which share no common divisor. Note that this implies that $a$, $b$, and $c$ are pairwise coprime. If $a$,$b$, and $c$ are a primitive Pythagorean triple satisfying the equation $a^2 + b^2 = c^2$ , is it then always true that

$$\gcd ( (a+b) , (a+b-c) ) = 1?$$

Can it be proven that $\gcd ( (a+b) , (a+b-c) ) = 1$ invariably ?

Can you find a counterexample where $\gcd ( (a+b) , (a+b-c) ) \neq 1$?

Answer 1

We have that $$ \gcd(a+b, a+b-c) = \gcd(a+b, -c) = \gcd(a+b, c). $$ Now suppose that $p$ is a prime dividing $\gcd(a+b, c)$. Because $a^2 + b^2 = c^2$, we have that $p^2$ also divides $(a+b)^2 - (a^2 + b^2) = 2ab$. In particular $p$ divides either $a$ or $b$ (even if $p=2$). But this contradicts our assumption.

Answer 2

Hint $\ (a\!+\!b, a\!+\!b\!-\!c)= (a\!+\!b,c)\mid (a\!+\!b,\!\!\!\overbrace{c^2}^{a^2\,+\,b^2}\!\!\!)\mid 2(a^2,b^2)\!=\!2,\,$ but $c$ is odd in a PPT