= Solution
Let $g$ generate the <cyclic group> of nonzero residues, whose order $p-1=2^{2^n}$ is a power of two. Write $a=g^j$. Its <multiplicative order> is $(p-1)/\gcd(j,p-1)$, so it is a <primitive root> precisely when $j$ is odd. The <quadratic residues> are exactly the even powers of $g$. Thus \b[the primitive roots are precisely the quadratic nonresidues], with nonzero <residue> classes understood.
For $n\geq1$, $p\equiv1\pmod4$, so <quadratic reciprocity> gives $(7/p)=(p/7)$. Since $2^3\equiv1\pmod7$ and $2^n$ alternates between $1$ and $2$ modulo three, $p$ is respectively $3$ or $5$ modulo seven. The nonzero squares modulo seven are $1,2,4$. Consequently $(7/p)=-1$, and
$$
\boxed{7\text{ is a primitive root modulo every Fermat prime }2^{2^n}+1\text{ with }n\geq1.}
$$
There is a genuine endpoint exception if the printed form permits $n=0$: then $p=3$ and $7\equiv1\pmod3$, which has order one and is not primitive. The first equivalence remains true for $p=3$, but the final assertion requires excluding that prime.
Back to article page
