Huh, I never realized this was this simple. Fleshing out the general case the post mentions. Suppose p=x^n-1 is prime. Note that: x^n - 1 = (x^n + x^(n-1)

Huh, I never realized this was this simple. Fleshing out the general case the post mentions. Suppose p=x^n-1 is prime. Note that: x^n - 1 = (x^n + x^(n-1) +... + x^2 + x) -( x^(n-1) +... + x^2 + x +1) =>x^n-1 = (x^(n-1) + x^(n-2) + ... + x + 1)(x-1) So if p=x^n-1 is prime, one of the factors must be equal to 1. If the first factor is 1, this implies x=0 and so the second factor is -1. Therefore the second factor must be 1, and x=2. Therefore if x^n-1 is prime, x must be 2. Damn that's cool. At first I thought any number 2^n-1 would be prime, but then I realized that obviously the left hand factor can have subfactors. Which is to say I remembered 15 exists.