Show that if $ p =2^n +1$ is a Fermat prime, then 3 is a primitive root modulo $p$

Question

I am not sure how to go about this problem. Does anyone have any insight into a good start/theorems I may need to employ?

Answer 1

HINT.-We assume $n\gt1$ in which case $3$ is not primitive root modulo $3=2^1+1$.

Suppose $1\le x, y\le2^n$ with $x\ne y$ and $3^x=3^y$. Suppose also that $x\gt y$.

We have $3^y(3^{x-y}-1)\equiv0\pmod{2^n+1}$. Because $2^n+1$ is prime greater than $3$, one has $3^y\ne0$ so $3^{x-y}-1=0\Rightarrow x-y=2^nk$ (by FLT).

Thus $3^x\ne3^y$ and all the $2^n$ elements $3,3^2,3^3,\cdots,3^{2n-1},3^{2n}=1$ are distinct so $3$ is primitive root.