Divisibility of $a^m + b^m$ by $a^n + b^n$ implies divisibility of m by n

Question

I'm working on a problem that states: Let $a, b, m, n$ be natural numbers, $a > 1$, and suppose that $a$ and $b$ have no common factors (GCD). We are asked to prove that if $a^m + b^m$ is divisible by $a^n + b^n$, then $m$ must be divisible by $n$.

I've tried using long division and modular arithmetic, but I'm stuck on how to proceed with the proof. The textbook solution simply suggests "carrying out long division," which I did without success.

After seeing the comments, one user pointed out that this problem $a,b,m,n$ be positive integers such that g.c.d.$(a,b)=1$ and $a^m+b^m\mid a^n+b^n$ , then $m\mid n$? offers the same solution.

And that is indeed similar. I have seen the answers, but I found them unhelpful. It assumed some prerequisite knowledge that I don't have, as of now, or perhaps, it skipped some steps or didn't give as elaborate of an explanation as my brain needed.

Okay, after seeing some comments, and answers, I have advanced my understanding of the problem.

So, I have assumed that $n <= m$, and I have used the division algorithm to express the following statement $m = nq + r$ for $0<=r<n$.

$a^n + b^n | a^m + b^m$. From this statement, I have thought of trying to do $a^m + b^m = (a^n + b^n)x$, and further try to solve the problem.

Can someone help me get started on the proof? Understanding the logic and intuition behind each step would be very helpful.

Answer 1

Here is a sketch: First show that $n\leq m$. Then note that $a^m+b^m=a^{m-n}(a^n+b^n)-a^{m-n}b^n+b^{m-n}(a^n+b^n)-b^{m-n}a^n$ $=(a^{m-n}+b^{m-n})(a^n+b^n)-a^{i}b^{i}(a^{j}+b^{j})$, where $i=\min(n,m-n)$ and $j$ accordingly either $m-2n$ or $2n-m$. The assumptions on $a,b$ imply that $a^n+b^n$ divides $a^{j}+b^{j}$. If $j=m-n$, then $j< m$ and if $j=2n-m$, then since $n\leq m$ we have $m\geq 2n-m$. In both cases $j\leq m$. In case of equality, $n$ divides $m$. If $j<m$ use induction, by which $n$ divides $j$ and thus $m$ as well.