Prove a sequence is a divisibility sequence

Question

Sequence: $a_n=2^n-1$

I want to prove that $a_m|a_n$ whenever $m|n$.

I started by approaching with induction, base/trivial cases being $m=n$, $n=0$, and $m=1$, but I'm not sure where to go from there or if there is a more straightforward method.

Note:

$m|n$ means for some $x$, $mx = n$

$a_m|a_n$ means for some $y$, $a_my=a_n$

Other similar questions appeal to laws beyond basic arithmetic and the definitions of divisibility and the sequence.

Use the fact that $a^n-b^n=(a-b)(a^{n-1} + a^{n-2}b + \ldots + b^{n-1})$ — ashK, Jan 30 '19 at 20:00

score 0 · Answer 1 · answered Jan 30 '19 at 20:00

0

If $m | n$ then $n = km$ so

$\begin{array}\\ 2^n-1 &=2^{km}-1\\ &=(2^{m})^k-1\\ &=(2^m-1)\sum_{j=0}^{k-1}2^{jm}\\ \end{array} $

answered Jan 30 '19 at 20:00

marty cohen

107,799

Prove a sequence is a divisibility sequence

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