I got this by watching a video on youtube and they said that this is always true but I am wondering why. I have tried to use Fermat's little theorem (FLT) but got nowhere bcs it says that if p is a prime then $p|{a^p-a}$.
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1Here are some of the first Google hits on this website when searching the title of this question (which is indicative of a good title, by the way): http://math.stackexchange.com/q/319963/, http://math.stackexchange.com/q/1045037/, http://math.stackexchange.com/q/186587/, http://math.stackexchange.com/q/763751/ – Jonas Meyer Jan 06 '15 at 14:37
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Side note: Your statement "$2^n-1\text{ is prime}\implies n\text{ is prime}$" can be extended to "$2^n-1\text{ is prime}\implies n\text{ is prime}\wedge(2^n-1)\cdot(2^{n-1})\text{ is perfect}$". – barak manos Jan 06 '15 at 14:49
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Use the contrapositive:
$2^n-1$ is composite if $n$ is composite
and observe that $2^{pq}-1$ is divisible by $2^p-1$.

MJD
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Some people use pictures of Georg Cantor or David Hilbert. I don't want to be so pretentious. Also, the potato is funny. – MJD Jan 06 '15 at 14:39
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Here's an elementary proof. Assume $n$ is not prime. Then $n=p \cdot q$ for some integers $p$ and $q$ such that $p \leq q$ and $p \neq 1$. Then $$2^n-1=(2^q)^p-1$$ factorizing that $$(2^q)^p-1^p=(2^q-1)((2^q)^{p-1}+ \cdots + 1)$$ Since $q \neq 1$, we have obtained factors for $2^n-1$ which shouldn't have been possible if it was prime, hence we have a contradiction.

sayantankhan
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