Show that if \[2^{m}+1\] is prime then
\[m=2^{n}\] \[n \in \mathbb{N} \]
Guidance: \[m=r*2^{n}\]
So I've started like this: \[2^{m}+1=2^{r*2^{n}} +1 = \left( 2^{2^{n}} \right)^{r}+1\]
I assume that's a somewhat correct beginning but how do I continue?

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have you shown why m has to be a multiple of \(2^n\)?

No, not really

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