Find the integers n such that 4^(n)-1 is prime.

4^(n)-1 factors to (2^(n)-1)(2^(n)+1) by difference of two squares by definition of a prime (only has factors 1 and itself (given that itself isn't 1))So Exactly one of the brackets should = 1 for 4^(n)-1 to be primeLooking at the first bracket, to see when it = 1, we see: (2^(n)-1) = 1 implies 2^(n) = 2 and so n = 1 is the solution. Checking the other bracket we see it equals 5 (which isn't 1). And so n = 1 is a solution.Now to make the 2nd bracket = 1 we use a similar method to see 2^(n) = 0 and we know that exponential functions never = 0 so there is no solution for the 2nd bracket = 1 .and so the only solution is n= 1. and to check that this is a solution we can substitute it in, giving 3, which is indeed prime.