How to prove that $\gcd(m^2,m+n) = \gcd(n^2,m+n)$ [duplicate]
How can we prove that $\gcd(m^2,m+n) = \gcd(n^2,m+n)$?
My try:
$$
\gcd(m^2,m+n) = \gcd\left[m^2=m\cdot \left(m+n\right)-mn\right] = \gcd(m+n,mn)$$
But I don't think it'll help me for the future, there is another way maybe?
As @WhatsUp said, you can continue this to a solution.
$$ \gcd(m+n,mn)=\gcd(m+n,-mn)$$ $$=\gcd(m+n,-mn+n(m+n))=\gcd(m+n,n^2)$$