elementary number theory - If $a,b inmathbb N$ and $gcd(a,b)=1$, prove that $gcd(a+b;a^2+b^2)= 1$ or $2$.

If $a,b \in\mathbb N$ and $\gcd(a,b)=1$, prove that $\gcd(a+b,a^2+b^2)$ is always equal to either 1 or 2, where $\gcd$ is the greatest common divisor. I haven't really ever solved a problem like this before, so I'd like to get some help. Thanks.

Answer

If integer $d$ divides both $a+b,a^2+b^2$

$d$ must divide $a^2+b^2+ (a-b)(a+b)=2a^2$

Similarly, $d$ must divide $a^2+b^2- (a-b)(a+b)=2b^2$

$\implies d$ must divide $(2a^2,2b^2)=2(a^2,b^2)$

As $(a,b)=1,(a^2,b^2)=1$

The GCD will be $1$ or $2$ according as $a,b$ are of opposite or same parity.