Let $a, m, n\in\mathbb{N^{*}}$ With $m>n$. Show ${(a^{2^m}-1,a^{2^n}+1)=a^{2^n}+1}$

Question

Let $a, m, n\in\mathbb{N^{*}}$ With $m>n$. Show $${{(a^{2^m}-1,a^{2^n}+1)=a^{2^n}+1}}$$$$$$My thoughts: If we could show that $$(a^{2^n}+1)\mid (a^{2^m}-1)$$ the property that gcd says "If $\;a\mid b\Longrightarrow (a,b)=a$" I can not work well with two unknowns in number theory.

Answer 1

Hint: $(x^2-1)=(x+1)(x-1)$, so $x-1|x^2-1$ and $x+1|x^2-1$.

Answer 2

First show that $(a^p-1,a^q-1) =a^{(p,q)}-1$ so that for $m>n$, $$(a^{2^m}-1,a^{2^{n+1}}-1)=a^{2^{n+1}}-1 = (a^{2^n}-1)(a^{2^n}+1)$$ from which it follows that $a^{2^n}+1$ is a divisor of $a^{2^m}-1$.