Let [itex]k\in\mathbb{N}[/itex], and [itex]q[/itex] be a prime factor of [itex]F_{k}=2^{2^{k}}+1[/itex].(adsbygoogle = window.adsbygoogle || []).push({});

Deduce that gcd[itex](q-1,2^{k+1})=2^{k+1}[/itex].

[itex]q|F_{k}[/itex] [itex] \Rightarrow [/itex] [itex] mq = 2^{2^{k}}+1[/itex] for some [itex]m\in\mathbb{N}[/itex]

[itex] 2^{2^{k}}=q-1+(m-1)q[/itex] [itex] \Rightarrow [/itex] [itex] 2^{2^{k}}=q-1[/itex] (mod [itex]q[/itex])

[itex]2^{k+1}|2^{2^{k}}[/itex] since [itex]k+1\leq 2^{k}, \forall k\in \mathbb{N}[/itex]

So [itex]2^{2^{k}}=n2^{k+1}[/itex] for some [itex]n\in \mathbb{N}[/itex].

I think I'm missing something, so any nudge in the right direction would be much appreciated.

Thanks

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# Fermat Numbers - Factor Form Proof

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