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Prove, when $n$ is a positive integer, if $2^n - 1$ is prime, then $n$ is prime.
I did read some sort of proving on the web, but I could not understand it...
And if possible, could the explanations be very clear?
Thank you guys :)
So, I've found this from somewhere:
Theorem. If 2n−1 is prime then n is prime.
Proof. Suppose that 2n−1 is prime, and write n=st where s,t are positive integers. Since xs−1=(x−1)(xs−1+xs−2+⋯+1) , we can substitute x=2t to see that 2t−1 is a factor of 2n−1. Since 2n−1 is prime there are only two possibilities, 2t−1=1or2t−1=2n−1 . Therefore t=1 or t=n. We have shown that the only possible factorisations of n are n×1 and 1×n. Hence, n is prime.
However, I have a few questions: 1) how do you know xs−1=(x−1)(xs−1+xs−2+⋯+1) 2) why is '2t−1' a factor of '2n−1' 3) how do you determine '2t−1=1or2t−1=2n−1' from 'Since 2n−1 is prime' 4) 'the only possible factorisations of n are n×1 and 1×n. ' how? 5) how did 'Hence, n is prime' come out as result?