Prove that the greatest common factor of $m+n$ and $m^2+n^2$ is 1 or 2 if $m$ and $n$ are relatively prime. Prove that the greatest common factor of $m+n$ and $m^2+n^2$ is $1$ or $2$ if $m$ and $n$ are relatively prime natural numbers.
Can anyone give a step-by-step answer for this?

Notice $\,{\rm mod}\ m\!+\!n\!:\,\ \color{#c00}{m\equiv -n}\,\Rightarrow\,f(\color{#c00}m)= m^2\!+n^2\equiv\, f(\color{#c00}{-n})\,\equiv\, \color{#0a0}{2n^2}$
By Euclid $\,\ (m\!+\!n,f(m)) = (m\!+\!n,\, f(m)\ {\rm mod}\ m\!+\!n) = (m\!+\!n, \color{#0a0}{2n^2})$
Furher $\ (m\!+\!n,n)=(m,n)=1\,\Rightarrow\, (m\!+\!n,n^2)=1\ \Rightarrow\ (m\!+\!n,2n^2) = (m\!+\!n,2)$
