For which Mersenne primes and Fermat primes is 2 a primitive root?

Question

Could you help me answering this question?

For which Mersenne primes and Fermat primes is 2 a primitive root?

I know that a $a$ is a primitive root modulo $n$ if $a$ generates $\mathbb{Z}_n^*$.

A Mersenne prime is a prime of the form $M_p=2^p-1$, with $p$ prime.

A Fermat prime is a prime of the form $2^{2^n}+1$, with $n\in \mathbb{N}$.

Thank a lot :)

Answer 1

Let $2^p - 1$ be a Mersenne prime and consider the subgroup of $\mathbb{F}_{2^p-1}^\ast$ generated by 2. It consists of $\{1, 2, 4, \ldots, 2^{p-1}\}$ since $$2^p \equiv 1 \pmod{2^p - 1}$$ and those elements are unique since each is less than $2^p - 1$. So that's a total of precisely $p$ elements in the subgroup $\langle 2 \rangle$.

So ask yourself: when is this the entirety of $\mathbb{F}_{2^p-1}^\ast$?

Answer 2

The order of $2\bmod 2^n-1$ is clearly $n$. The order of the group however is $2^n-2$. So we need $n=2^n-2$. Clearly the only solution is $n=2$. So The Mersenne prime needed is $3$.

The order of $2\bmod 2^{k}+1$ is $2k$. The order of the group however is $2^k$. So we need $2k=2^k$. The only solution is clearly $k=2$. so the fermat prime needed is $5$.

Answer 3

hint

For $n \geq 2$, $F_n \equiv 1 \pmod{8}$, hence $2$ will be a quadratic residue so cannot be a primitive root.

For $p \geq 3$, $2^p \equiv 1\pmod{M_p}$, thus the order of $2$ modulo $M_p$ is at most $p$ so it cannot be a primitive root.