Prove that $(p^m-1) \mid (p^n-1) \Leftrightarrow m \mid n$. The $\Leftarrow$ part is ok, the $\Rightarrow$ part should be easy but I'm stuck with it! Thanks in advance.
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I started writing $p^n-1=a(p^m-1)$ but I wasn't able to arrive at the conclusion I want... – PITTALUGA Nov 10 '13 at 12:51
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Related : http://math.stackexchange.com/questions/7473/prove-that-gcdan-1-am-1-a-gcdn-m-1 – lab bhattacharjee Nov 10 '13 at 13:07
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OK I got it, thanks for all your suggestions! – PITTALUGA Nov 10 '13 at 13:43
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1possible duplicate of Showing that $a^n - 1 \mid a^m - 1 \iff n \mid m$ – Bart Michels Apr 15 '15 at 08:06
3 Answers
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This follows from the fact that, given integers $a>b>0$, then $(a^n-b^n)_{n\ge 1}$ is a divisibility sequence (see also this article on Arxiv).
Paolo Leonetti
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We are required to prove:
If $p^m-1\mid p^n-1$, then $m\mid n$
We can state this in an equivalent contrapositive
If $m\nmid n$, then $p^m-1\nmid p^n-1$
We can prove the second statement by contradiction:
Assume there exist integers $m, n$ such that $m\nmid n$, but $p^m - 1 \mid p^n - 1$. How will you prove this?
Hanul Jeon
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Gerard
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How is this not just restating the problem? This doesn't seem to provide guidance in how to take a step forward... – Steven Stadnicki Nov 11 '13 at 03:36
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It does. In a mathematical sense, everything is just restating the problem. Additionally, a lot of theorems have been proved by actually proving their contrapositives. Take the Steiner-Lehmus problem, for example. – Gerard Nov 11 '13 at 10:10