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Trying out a few examples, let (i) $a=3, n=1, m=2, 3^2+1=10, 3^4-1=80, 10\mid 80$, (ii) $a=3, n=2, m=3, 3^4+1=82, 3^8-1=6560, \frac{6560}{82}= 80$.

To derive a proof, should derive $a^{2^n} +1$ as a factor of $a^{2^m} -1$, & as $m\gt n$, so $a^{2^m}=a^{2^n}.a^{2^{(n-m)}}$, so $a^{2^m}-1=a^{2^n}.a^{2^{(n-m)}}-1$.

It is possible to state the next step, based on something like $a^2-1=(a-1)(a+1)$, but lack familiarity with such type of polynomial (if chosen the correct word) division.

Also, would request a generalized answer, for different bases for powers of $n,m$.

jitender
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    Start with $,a^{2^m}-1=\left(a^{2^{m-1}}-1\right)\left(a^{2^{m-1}}+1\right),$, then repeat as necessary. – dxiv Mar 01 '18 at 00:40
  • @dxiv That is seemingly an inductive approach. – jitender Mar 01 '18 at 00:42
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    @jitender Right. It can also be written as a telescoping product, which "hides" the same idea of induction. – dxiv Mar 01 '18 at 00:44
  • @dxiv Sorry, not comfortable with telescoping product. If you could please elaborate that in answer to this post. – jitender Mar 01 '18 at 00:45

2 Answers2

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Hints (either of the following):

  • $a^{2^{m}} - 1 = \left(a^{2^{m-1}} - 1\right)\left(a^{2^{m-1}} + 1\right)$ , so $\,a^{2^{m-1}} - 1 \mid a^{2^m} - 1 \,$, then $\,a^{2^{n}} - 1 \mid a^{2^m} - 1 \,$ for $\,n \lt m\,$ by induction

  • let $\,N=\left(a^{2^{n}} + 1\right)\left(a^{2^{n+1}} + 1\right)\ldots\left(a^{2^{m-1}} + 1\right)\,$, then telescoping $\,(x-1)(x+1)=x^2-1\,$:

$$ \begin{align} \left(a^{2^{n}} - 1\right)N &= \left(a^{2^{n}} - 1\right)\left(a^{2^{n}} + 1\right)\left(a^{2^{n+1}} + 1\right)\ldots\left(a^{2^{m-1}} + 1\right) \\ &= \left(a^{2^{n+1}} - 1\right)\left(a^{2^{n+1}} + 1\right)\ldots\left(a^{2^{m-1}} + 1\right) \\ &= \cdots \\ &= \left(a^{2^{m-1}} - 1\right)\left(a^{2^{m-1}} + 1\right) \\ &= \left(a^{2^{m}} - 1\right) \end{align} $$

dxiv
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\begin{eqnarray*} a^{2^{m}}-1 &=& (a^{2^{m-1}}+1)(a^{2^{m-1}}-1) \\ &=& (a^{2^{m-1}}+1)(a^{2^{m-2}}+1)(a^{2^{m-2}}-1) \\ & & \vdots \\ &=& (a^{2^{m-1}}+1)(a^{2^{m-2}}+1)\cdots \color{red}{(a^{2^{n}}+1)}\cdots (a^2+1)(a+1)(a-1). \end{eqnarray*}

Donald Splutterwit
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