Show that if a^n - 1 is prime then a = 2. If n is not prime, can 2^n-1 be prime?
This question is based on MAT 2015 question 2. This question has already built up the generalized difference of two powers in the buildup. We can apply this to an - 1 to give: an - 1 = (a - 1)(an-1 + an-2 + ... + a + 1).From this we can see that an - 1 is always divisible by a - 1. So for an-1 to be prime we must have that a-1 = 1, so a = 2.Now that we have a = 2, we can change future statements about an-1 to 2n-1. Now assume that n is not prime, that is n is composite. Then n = pq for some p,q > 1. Now we can rewrite:2n - 1 = 2pq - 1 = ( 2p)q - 1 = (2p-1)( (2p)q-1 + (2p)q-2 + ... + 2p + 1)Since p > 1, we have that 2p - 1 > 21 - 1 = 1. Therefore 2n-1 is divisible by 2p-1 and is not prime.
