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

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!

Solution 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 $x^k-1=(x-1).(\ldots)$; 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 $(a^{2^{n+1}})^{2^{m-(n+1)}} = a^{2^{n+1}.2^{m-(n+1)}} = a^{2^m}$.)

Solution 2:

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