Prove that $gcd(a+b,a^2+b^2) \le 2$

Question

Let $a$ and $b$ be two integers with $ gcd(a,b) = 1$

Prove that $gcd(a+b,a^2+b^2) \le 2$

Answer 1

If a prime $p$ divides $a+b$ and $a^2+b^2$ it also divides

$$(a+b)^2 - (a^2+b^2) = 2ab$$

and so it divides $2$ (so $p=2$), or $p|a$ or $p|b$. But if $p|a$ then $p|b$ etc. (as $p | (a+b) \land p|a \to p|b$), so the second case or third case cannot happen, as $a$ and $b$ have no common divisor. So $p=2$ remains, and this happens e.g. when $a=1,b=3$, as an example.

So the gcd can only be a power of $2$, so if we can rule out $4$ as a common divisor we're done. But the sum of two squares $a^2 + b^2$ can only be divisible by $4$ if both $a$ and $b$ are even (modulo $4$ squares are $0$ or $1$), and this cannot happen, again as $a$ and $b$ are relatively prime.