<s>[INST] If $d=\gcd(a+b,a^2+b^2)$, with $\gcd(a,b)=1$, then $d=1$ or $2$ Suppose $\gcd(a,b)=1$. Let  $d=\gcd(a+b,a^2+b^2)$.  I want to prove that $d$ equals $1$ or $2$.
I get that $d\mid2ab$ but I can't find a linear combination that will give me some help to use the fact that $\gcd(a,b)=1$.
 [/INST] Bezout based approach
Since $(a,b)=1$, there are $x,y$ so that $ax+by=1$. Then
$$
\begin{align}
1
&=(ax+by)^3\\
&=a^2\color{#C00000}{(ax^3+3bx^2y)}+b^2\color{#00A000}{(3axy^2+by^3)}
\end{align}
$$
Then, since
$$
(a^2+b^2)+(a+b)(a-b)=2a^2
$$
and
$$
(a^2+b^2)-(a+b)(a-b)=2b^2
$$
we have
$$
\begin{align}
&(a^2+b^2)\left[\color{#C00000}{(ax^3+3bx^2y)}+\color{#00A000}{(3axy^2+by^3)}\right]\\
+&(a+b)(a-b)\left[\color{#C00000}{(ax^3+3bx^2y)}-\color{#00A000}{(3axy^2+by^3)}\right]\\
=&2a^2\color{#C00000}{(ax^3+3bx^2y)}+2b^2\color{#00A000}{(3axy^2+by^3)}\\[6pt]
=&2
\end{align}
$$
Therefore, $(a+b,a^2+b^2)\mid2$.
</s>