A different way to show that $a^k-1 \mid a^n-1$ if $k\mid n$?

Question

A student that I'm tutoring showed me the following problem:

Let $a\gt 1,$ and $k,n\gt 0.$ Prove that $a^k-1\mid a^n-1$ if and only if $k\mid n$.

Solution: Since $k$ divides $n$, we have $n=ks$ for some integer $s$. $a^n-1=a^{ks}-1=(a^k)^s-1=(a^k-1)((a^k)^{s-1}+(a^k)^{s-2}+\dots+(a^k)^2+a^k+1)$. Thus, $a^k-1$ divides $a^n-1$.

Now suppose that $a^k-1$ divides $a^n-1$. Let $n=ks+r$ where $r\ (0\le r\le k-1)$ is a remainder when we divide $n$ by $k$. $a^n-1=a^{ks}a^r-1=((a^k)^s-1)a^r+(a^r-1)$. By the argument above, $a^k-1$ divides $a^{ks}-1.$ Thus, $a^k-1$ must divide $a^r-1$. Since $r\lt k,$ we have $r=0.$

He wanted to see another way to prove the converse and I got stumped! Can anyone think of an easy proof of why $a^k-1 \mid a^n-1$ if $k\mid n$?

Answer 1

A modular way: $\ {\rm mod}\ a^{\large k}\!-\!1\!:\,\ \color{#c00}{a^{\large k}\equiv 1}\,\Rightarrow\, a^{\large r+kq}\equiv a^{\large r}(\color{#c00}{a^{\large k}})^{\large q}\equiv a^{\large r}\color{#c00}1^{\large q}\equiv a^{\large r}\,$

This uses standard Congruence Rules.

In the same way $\, b\equiv 1\,\Rightarrow\, f(b)\equiv f(1)\,$ for any polynomial $\,f(x)\,$ with integer coefficients (above is $\,f(x) = x^q,\ b = a^k).$ So the result, is a special case of the Factor Theorem $\,b\!-\!1\mid f(b)\!-\!f(1),\,$ which is a special case of the linked Polynomial Congruence Rule. The advantage of the modular method is that one doesn't need to compute any quotients (only remainders, or congruent numbers).

Remark $\ $ One can prove further that $\, (a^k\!-\!1,a^n\!-\!1) = a^{(n,k)}\!-\!1\,$ where $\,(x,y):=\gcd(x,y).\,$ This is a special case of a strong divisibility sequence.