How to show $a^{2^n}+1 \mid a^{2^m}-1$?

Question

I've been struggling with this all day today. I imagine it's not very hard, but my algebra skills are terrible. So, how can I show that if $m>n$ and $a$ is a positive integer, then $$a^{2^n}+1 \mid a^{2^m}-1.$$ I just can't get a coherent picture of what I'm supposed to do. If this too (see yesterday's question) boils down to the Unique Factorization Theorem, I would be very grateful for any intuition on how to think about this!

Answer 1

First you could try to show that: $x-1\mid x^k-1$ for any integer $x$ and any positive integer $k\ge 1$. (This follows from the well-known equality $x^k-1=(x-1)\cdot(\ldots)$; try to fill in the dots.)

Now you can use:

$(a^{2^n}-1)(a^{2^n}+1)=(a^{2^n})^2-1=a^{2^{n+1}}-1$

This means that $a^{2^n}+1 \mid a^{2^{n+1}}-1$ and using the above result for $x=a^{2^{n+1}}$ and $k=2^{m-(n+1)}$ you get $a^{2^{n+1}}-1\mid a^{2^m}-1$.

(Note that $(a^{2^{n+1}})^{2^{m-(n+1)}} = a^{2^{n+1}.2^{m-(n+1)}} = a^{2^m}$.)

Answer 2

$$(a^{2^n}+1)(a^{2^n}-1)(a^{2^{n+1}}+1)(a^{2^{n+2}}+1)\cdots(a^{2^{m-1}}+1) = a^{2^m}-1$$

Answer 3

Hint $ $ Below put $\rm\ X = a^{\large 2^{\Large N}}, \ 2K = 2^{M-N}\ $ (cf. coprimality of Fermat numbers)

The Factor Theorem $\rm\, \Rightarrow\, X+1\mid X^{\large2K}-1\ $ since $\rm\ X = -1\ $ is a root of the latter.

Alternatively $\ \rm\bmod\ X+1\!:\ \ \color{#c00}{X\equiv -1}\, \Rightarrow\ \color{#c00}X^{\large 2K}\!\equiv (\color{#c00}{-1})^{\large2K}\!\equiv 1\, $ via Congruence Power Rule.

Note in particular how the use of modular arithmetic enables one to reduce the proof to the triviality that $\rm\ (-1)^2 = 1.\,$ Such order $\,2\,$ cyclicity is ubiquitous, e.g. the test for divisibility by $11,\, $ where $\rm\ 10\equiv -1\ \Rightarrow\ 10^{\large 2K}\equiv 1,\:\ 10^{2K+1}\equiv -1.$