
#1
Oct2209, 07:00 AM

P: 74

I was trying to prove the statement "If (2^n)1 is prime then n is prime". I've already seen the proof using factorisation of the difference of integers and getting a contradiction, but I was trying to use groups instead. I was wondering if it's possible, since I keep getting stuck.
So far I've got: If (2^n)1=p where p is prime then [tex]2^{n}\equiv1 mod(p)[/tex]. The group {1,2,...,p1} is cyclic and every element has order p1. So n = k(p1) for some positive integer k. But doesn't this mean n is not prime? which is wrong I know. Could someone point me in the right direction or a theorem that might be useful? 



#2
Oct2309, 11:56 PM

P: 105

Interesting problem... I'm not yet sure of a proof, but your statement



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