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.

Question

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?

Answer 1

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 $\ (m\!+\!n,n)=(m,n)=1\,\Rightarrow\, (m\!+\!n,n^2)=1\ \Rightarrow\ (m\!+\!n,2n^2) = (m\!+\!n,2)$

Answer 2

If prime $p$ divides both, $p$ divides $(m+n)^2-(m^2+n^2)=2mn$

If $p$ divides $m,p$ must divide $m+n-m=n$

$p$ must divide $(m,n)=1$

Answer 3

Let $d\lvert m+n$ and $d\lvert m^2+n^2$.

Then $d\lvert(m^2+n^2)-(m-n)(m+n)$, so $d\lvert 2n^2$;

$\;$and $d\lvert(m^2+n^2)+(m-n)(m+n)$, so $d\lvert 2m^2$.

Since $\gcd(2m^2, 2n^2)=2\gcd(m^2,n^2)=2\cdot1=2$,

$\hspace{.3 in}d\lvert2$ so $d=1$ or $d=2$.