About Primitive roots

Question

Good day! I'm currently studying on the primitive roots mod n. Eventually, I fully understand the concept of calculating the primitive roots of a number by practice, but I encounter the following problems that is out of my league. Please help me understand the following concept. Any idea will be of great help.

1. Show that $$ Fn = 2 ^ {2 ^ n} + 1 $$ , n > 1, is a prime, then 2 is not a primitive roots of Fn.

2. Show that if p = 1 (mod 4) and $g$ is a primitive root of $p$, then so is $-g$. Show by a numerical example that this need not be the case if p = 3 (mod 4).

3. Prove that if $a$ has order hk (mod m), then $a^{k}$ has order k mod n.

4. Prove that if a has order 2k modulo p the odd prime p, then a^k = -1 (mod p).

Answer 1

(3) We show the correct fact: $h$ is the order of $a^k$ modulo $n$.

Since $a$ has order $hk$ modulo $m$, we have $a^{hk} \cong 1 \pmod{n}$ and $hk$ is the least positive power of $a$ congruent to $1$. The powers of $a^k$ are a subsequence of the powers of $a$ and $a^{hk} = (a^k)^h$ is the smallest positive power of $a$ that is congruent to $1$ and so $h$ is the smallest power of $a^k$ that is congruent to $1$. Therefore, $h$ is the order of $a^k$ modulo $n$.

Answer 2

Here is a solution to the first question. (Maybe ill add more as I see fit, the others are easier.) Let $p=2^{2^n}+1$ be prime. Then $$2^{2^n}\equiv -1 \mod p$$ so

$$2^{2^{n+1}}\equiv 1 \mod p$$ Now if $2$ were a primitive root its order would be $p-1$ and so $p-1|2^{n+1}$ or $2^{2^n}|2^{n+1}$ which means that $2^n\leq n+1$ which is false for $n>1$. Thus $2$ is not a primitive root.