<s>[INST] Show that $\gcd(a + b, a^2 + b^2) = 1\mbox{ or } 2$ How to show that $\gcd(a + b, a^2 + b^2) = 1\mbox{ or } 2$ for coprime $a$ and $b$?
I know the fact that $\gcd(a,b)=1$ implies $\gcd(a,b^2)=1$ and $\gcd(a^2,b)=1$, but how do I apply this to that?
 [/INST] Let positive integer $d$ divides both $a+b,a^2+b^2$
$\implies d$ divides $(a^2+b^2)+(a+b)(a-b)=2a^2$ 
Similarly, i.e., $d$  divides  $(a^2+b^2)-(a+b)(a-b)=2b^2$
$\implies d$ divides $2a^2,2b^2\implies d$ divides $(2a^2,2b^2)=2(a^2,b^2)=2(a,b)^2$
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