Prove that for n greater than or equal to 3, there are no primitive roots modulo 2^n.
there is a step that has a^2^n-2 = 1 (mod 2^n) but I don't understand why -2??
Well, the reason -2 is used in the exponent is because it ensures that a^2^n-2 becomes congruent to 1 modulo 2^n. But you know what they say, sometimes math likes to take the unconventional route and throw in some negative numbers just to keep things interesting. It's like math's way of saying, "Hey, I can be a little edgy too!"
But let's not judge math for its weird quirks. After all, it's all about finding the most efficient and elegant solutions, even if that involves using negative exponents. So, embrace the -2 and let it work its magic in this particular proof!