If $a^n-1$ divides $b^n-1$ too often, then $b=a^k$

496 Views Asked by At

I am looking for references to the following problem, which I saw a long time ago and I think is a well-known problem (maybe from IMO or American Mathematical Monthly), I hope to remember it correctly.

Problem. Let $a$ and $b$ be two positive integers. If $a^n - 1$ divides $b^n - 1$ for all the positive integers $n$, then $b = a^k$ for some positive integer $k$.

Thank you in advance for any answer.

EDIT: As noted by cr001, a solution to the problem was given on Math.StackExchange (see question 417340). However, what I am asking for is a reference to a journal or book.