Question 409839
I sure hope you mean {{{2^n - 1}}} instead of {{{2n-1}}}. In fact I am working on a mathematics project regarding odd perfect numbers, and such a prime in the form {{{2^n - 1}}} is important since there is a one-to-one correlation between these primes (called Mersenne primes) and the *even* perfect numbers.


Suppose that n is not prime, i.e. n = ab for positive integers a and b, where a and b are not equal to 1. Then, {{{2^n - 1 = 2^ab - 1}}}, and this can be factored as {{{(2^a - 1)*sum(2^ai, i = 1, b-1)}}} (this comes from the geometric series). Since neither factor is 1, then {{{2^n - 1}}} can be factored as a product of two integers, hence it is not prime. The contrapositive is true, i.e. if {{{2^n - 1}}} is prime, then {{{n}}} is prime. The converse is not necessarily true, and we can find counterexamples, such as n = 11, as {{{2^11 - 1 = 23*89}}}.


Part b is a little tricky, I spent a bit of time on the problem and couldn't come with a full solution. Perhaps you could use an inductive argument, or show that {{{2^(2^k) + 1}}} cannot be factored for any k (which might also use induction).