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I started with $n = 2k$ , $2^{2k} - 1 = (2^k - 1)(2^k + 1)$, where i go from there...

JoeSuli
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2 Answers2

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Welcome to MSE!

Hint:

If $n = kl$ shows $n$ is composite, then we can write

$$ 2^n - 1 = 2^{kl} - 1 = (2^k)^l - 1^l $$

But do you know of a way to factor $a^l - b^l$?


I hope this helps ^_^

Brian M. Scott
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HallaSurvivor
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$ n $ not prime $ \implies $

$$\exists p,q>1 \;\;:\;\; n=pq\implies$$

$$2^n-1=(2^p)^q-1$$ $$=(2^p-1)(1+2^p+2^{2p}+...+2^{(q-1)p})$$

is not prime.