How can I prove that $a^m-1|a^n-1 \Leftrightarrow m|n$?

Question

Suppose that $(a,m,n)\in \mathbb{N}^{3}$, as $a>1$

I proved that $m|n \implies a^m-1|a^n-1$, but I couldn't prove the reciproque.

http://math.stackexchange.com/questions/7473/prove-that-gcdan-1-am-1-a-gcdn-m-1 — lab bhattacharjee, Mar 08 '14 at 10:21
See http://math.stackexchange.com/questions/486626/if-am-1-mid-an-1-then-m-mid-n and http://math.stackexchange.com/questions/413473/showing-that-an-1-am-1-iff-n-m — Martin Sleziak, Mar 08 '14 at 12:19

score 3 · Accepted Answer · answered Mar 08 '14 at 10:29

Suppose $a^m-1\mid a^n-1$. I claim $m\mid n$. Write $n=qm+r$ for $0\leq r<m$. Then notice $$ a^n-1=(a^{qm+r}-a^r)+(a^r-1)=a^r(a^{qm}-1)+(a^r-1). $$ But $m\mid qm$, so by your previous work, you know $a^m-1\mid a^{qm}-1$. This implies $a^m-1\mid a^r-1$. But $a^m-1>a^r-1$ since $m>r$ and $a>1$, so necessarily $a^r-1=0$, or $r=0$. Thus $m\mid n$.

score 1 · Answer 2 · answered Mar 08 '14 at 11:10

From different perspective, we can show by group theory.

it is enough to know that roots of $x^k-1$ is a group with $k$ elements.

suppose $x^m-1$ | $x^n-1$ that means that every root of first polynomial is also a root of second polynomial.

That means a group of $n$ elements include a group of $m$ elements by lagrange theorem; $m$| $n$.

How can I prove that $a^m-1|a^n-1 \Leftrightarrow m|n$?

2 Answers2