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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.

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

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  • $\begingroup$ This I propose to do by induction? In $m$ or $n$? $\endgroup$ – marcelolpjunior Aug 14 '13 at 19:48
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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 $a^{2^m}-1$.

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