elementary number theory - Prove $gcd(a+b,a^2+b^2)$ is $1$ or $2$ if $gcd(a,b) = 1$

Assuming that $\gcd(a,b) = 1$, prove that $\gcd(a+b,a^2+b^2) = 1$ or $2$.

I tried this problem and ended up with
$$d\mid 2a^2,\quad d\mid 2b^2$$
where $d = \gcd(a+b,a^2+b^2)$, but then I am stuck; by these two conclusions how can I conclude $d=1$ or $2$?
And also is there any other way of proving this result?

Answer

From what you have found, you can conclude easily.

If $d$ divides two numbers, it also divides their gcd, so

$$d| \gcd (2a^2,2b^2) = 2 \gcd (a,b) ^2 =2.$$

So, $d$ is a divisor of 2 and thus either 1 or 2.

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