Prove that $(a^m+b^m)\mid(a^n+b^n) \iff m\mid n$. Here $a, b, m, n\in\mathbb{Z}^+$, $m\leq n$ and $(a, b)=1$.
This is a questions from a number theory book that I am recently studying.
I have read through the solutions of $(a^m+1)\mid(a^n+1)$ and $(a^m-1)\mid(a^n-1)$. However, I wish to show the stronger statement of $(a^m+b^m)\mid(a^n+b^n)\iff m\mid n$.
My attempts:
When $n\geq 2m$, $(a^n+b^n)=(a^m+b^m)(a^{n-m}+b^{n-m})-a^mb^m(a^{n-2m}+b^{n-2m})$. Hence, $(a^m+b^m)\mid a^mb^m(a^{n-2m}+b^{n-2m})$. However, since $(a, b)=1$, $(a^m+b^m, a^mb^m)=1$, so $(a^m+b^m)\mid(a^{n-2m}+b^{n-2m})$ and we get another equation of the form $(a^m+b^m)\mid(a^n+b^n)$. We can repeat this process until $n<2m$.
Hence, suppose that $n<2m$. It suffices to prove that $n=m$. This is where I got stuck. How do I prove that $n=m$? Any help will be appreciated.
