<s>[INST] 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. 
 [/INST] Since $\gcd(a,b)=1$, Bezout's Identity says we have an $x$ and $y$ so that
$$
ax+by=1\tag{1}
$$
Note that
$$
\begin{align}
2a^2&=(a^2+b^2)+(a+b)(a-b)\\
2ab&=(a+b)^2-(a^2+b^2)\\
2b^2&=(a^2+b^2)-(a+b)(a-b)\\
\end{align}\tag{2}
$$
Therefore, incorporating $(1)$ and $(2)$,
$$
\begin{align}
2
&=2(ax+by)^2\\
&=2a^2x^2+4abxy+2b^2y^2\\
&=\Big((a^2+b^2)+(a+b)(a-b)\Big)x^2\\
&+2\Big((a+b)^2-(a^2+b^2)\Big)xy\\
&+\Big((a^2+b^2)-(a+b)(a-b)\Big)y^2\\
&=\color{#00A000}{(x-y)^2}\color{#C00000}{(a^2+b^2)}
+\color{#00A000}{((x^2-y^2)(a-b)+2xy(a+b))}\color{#C00000}{(a+b)}\tag{3}
\end{align}
$$
Equation $(3)$ says that
$$
\gcd(a+b,a^2+b^2)\,|\,2\tag{4}
$$
Note that $\gcd(1+2,1^2+2^2)=1$ and $\gcd(1+3,1^2+3^2)=2$, so both $1$ and $2$ are possible.
</s>