The following is exercise 6 of Chapter 4 in Ireland and Rosen's Number Theory.
If $p=2^n+1$ is a Fermat prime, show that $3$ is a primitive root modulo $p$.
I first recall that any Fermat prime actually has form $2^{2^n}+1$. Hence $p\equiv 1\pmod{4}$. Exercise 4 from the same chapter states the if $p\equiv 1\pmod{4}$, then $a$ is a primitive root mod $p$ iff $-a$ is as well. I was able to prove this, but unable to show $-3$ is a primitive root.
Is there a fruitful approach?
P.S. I was able to cheat and use the fact that $ \phi(p-1)=\phi(2^{2^n})=2^{2^n-1}=(p-1)/2 $ and since there are $(p-1)/2$ quadratic nonresides and each of the $\phi(p-1)$ primitive roots is a quadratic nonresidue, then the two sets are actually the same.
Then $3$ is a nonresidue since applying quadratic reciprocity $ \left(\frac{3}{p}\right)=\left(\frac{p}{3}\right)=\left(\frac{2}{3}\right)=-1 $ hence not primitive. But I only know this from a number theory class I took back in school, not from anything in Ireland and Rosen so far. Is there a way to avoid using a sledgehammer the authors haven't given me yet?