If $a^n-1$ is divisible by $b^n-1$ for all $n$, then $a$ is a power of $b$

Question

Let $a,b$ be natural numbers not equal to $1$ such that $\frac{a^n-1}{b^n-1}$ is natural for any natural $n$. Prove that $a=b^m$ for some natural $m$.

Answer 1

for all $b\neq 1,$ just write $$b^{mn}-1=(b^n-1)(1+ b+ b^2+....+b^{mn-n-1})$$ From here, it should be easy to get your answer.

Answer 2

You can find the solution here http://www.artofproblemsolving.com/Forum/viewtopic.php?f=59&t=4556