Prove that any positive integer that divides both $c-a$ and $c+a$ must be equal to 1 or 2

Question

Consider $b^2=c^2-a^2=(c-a)(c+a)$. Prove that any positive integer that divides both $c-a$ and $c+a$ must be equal to 1 or 2 . in particular this means that $GCD(\frac{c-a}{2},\frac{c+a}{2})=1$

How to prove this can any help

Answer 1

This is not true. If $a=6$ and $c=10$, then $4$ divides both $c-a$ and $c+a$.