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I've been struggling with this problem for a while now:

Prove that, given positive integers $m$ and $n$, if $m | n$ then $2^m-1 | 2^n-1$.

I can't seem to get any traction with this problem. Most of my approaches to the problem depend on knowing something about $\gcd(m,n)$, and all I know from the problem is that $\gcd(m,n) \neq 1$ when $m,n \neq 1$.

Can anyone give me a hint or suggest an approach I'm not seeing?

Thanks

Jim
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    This has surely been asked and answered here many times. The closest I could find, though, is http://math.stackexchange.com/questions/561077/pm-1-mid-pn-1-leftrightarrow-m-mid-n – The Chaz 2.0 Nov 13 '13 at 02:54
  • @TheChaz2.0 Ah, geez. Sorry I couldn't find anything in my search. Thanks for the link. – Jim Nov 13 '13 at 02:55
  • @Jim: if I had any clue how to search math-y stuff on here, I could have provided many more (and more helpful)links! – The Chaz 2.0 Nov 13 '13 at 02:57
  • You don't seem to be thinking about this the right way: You say that your approach depends on knowing about $\mathrm{gcd}(m,n)$. In this case, you know a lot about this number: If $m\mid n$, then $\mathrm{gcd}(m,n)=m$. – Andrés E. Caicedo Nov 13 '13 at 02:57
  • Granted, google gave me this: http://math.stackexchange.com/questions/529111/why-if-n-mid-m-then-an-1-mid-am-1 – The Chaz 2.0 Nov 13 '13 at 02:59
  • http://math.stackexchange.com/questions/7473/prove-that-gcdan-1-am-1-a-gcdn-m-1 – lab bhattacharjee Nov 13 '13 at 03:09

3 Answers3

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If $m\vert n$ then $2^n-1 = 2^{mk} - 1 = (2^m)^k -1 = \left(2^m - 1\right)\left((2^m)^{k-1} + (2^m)^{k-2} + \cdots +1\right)$

so $2^m -1\vert 2^n-1$

user85798
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Note that $2^m \equiv 1\pmod{2^m-1}$. If $m$ divides $n$, then $n=km$ for some positive integer $k$. It follows that $$2^n =2^{mk}=(2^m)^k \equiv 1^k=1\pmod{2^m-1}.$$

André Nicolas
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The other answers are better than this, in that they help learn concepts that are important, but: it's pretty easy to prove this by induction on $n/m$, since $(2^{n+m}-1)-(2^n-1)$ is easily seen to be a multiple of $2^m-1$.

Greg Martin
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