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I'm pondering the following number theoretical problem:

Show that if $a^m-1 \mid a^n -1$ then $m \mid n$.

My first idea was to use the factorization $a^m -1 = (a-1)(a^{m-1}+a^{m-2}+\cdots+a+1)$ in the condition $d(a^m-1)=a^n-1$, but I can't seem to bring anything out of that.

My second idea was to consider the cases where $a^m-1$ and $a^n-1$ are even/odd, using $a^m-1=2k$ or $2k+1$, respectively. But that doesn't seem to bring about anything either.

I've tried WLOG to assume $m \leq n$ and factor the divisibility condition as $da^m -d = a^n-1 \Leftrightarrow a^m(d-a^{n-m}) = d-1$, but that doesn't seem to do anything for me either.

Any ideas?

Numbersandsoon
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2 Answers2

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Hint Write $n=qm+r$ with $0 \leq r \leq m-1$.

Then

$$a^n-1=a^n-a^r+a^r-1$$

Try to prove that $a^m-1$ always divides $a^{qm+r}-a^r$.

From here you get that $a^m-1|a^n-1$ if and only if $a^m-1 | a^r-1$. Since $0 \leq r \leq m-1$ it is easy to finish the problem.

N. S.
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You can use a slow version of the division algorithm to do this.

Let $d=a^m-1$. We note that $m \leq n$ (a larger number cannot be a factor of a smaller one, and the hypothesis is that $d|a^n-1$.

From this we deduce that $d|(a^n-1)-(a^m-1)=a^m(a^{n-m}-1)$ and so $d|a^{n-m}-1$ (because $(d,a^m)=1$).

Now $n=mq+r$ with $0\leq r \lt m$, and repeating the process of subtracting $d$ and cancelling a factor if $a^m$ as often as necessary we find $d|a^r-1$. Since $r \lt m$ the only possibility is that $r=0$, so $n=mq$.

Mark Bennet
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