How can I find the possible values that $\gcd(a+b,a^2+b^2)$ can take, if $\gcd(a,b)=1$ If $\gcd(a,b)=1$, how can I find the values that $\gcd(a+b,a^2+b^2)$ can possibly take? I can't find a way to use any of the elemental divisibility and gcd theorems to find them. 

$$ \gcd(a+b,a^2+b^2) \mid \gcd((a+b)(a-b), a^2+b^2) = \gcd(a^2-b^2, a^2+b^2) \mid \gcd [ ( a^2+b^2)+ (a^2-b^2) , ( a^2+b^2)+ (a^2-b^2) ]=2 \gcd(a^2,b^2)=2$$
Now it is easy to check that both 1 and 2 are possible...
