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)(a^n-1 + a^n-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 = 2^pq - 1 = ( 2^p)^q - 1 = (2^p-1)( (2^p)^q-1 + (2^p)^q-2 + ... + 2^p + 1)Since p > 1, we have that 2^p - 1 > 2¹ - 1 = 1. Therefore 2ⁿ-1 is divisible by 2^p-1 and is not prime.