Is it true that :
$17$ is primitive root modulo $F_n(6)$
where $F_n(6)$ is generalized Fermat prime of the form:
$F_n(6) =6^{2^n}+1 , ~~ n \geq 0$
I know that one can use quadratic reciprocity to show that $a$ is primitive root modulo $p$ .
For example :
$F_n(20)=20^{2^n}+1 \equiv (-1)^{2^n}+1 \equiv 2 \pmod 3$
and we know that. $F_n(20) \equiv 1 \pmod 4~~$ therefore :
$\left( \frac{3}{F_n(20)}\right)=\left( \frac{F_n(20)}{3}\right)=\left(\frac{2}{3}\right)=-1$
so , $~~3~~$ is quadratic nonresidue mod $~~F_n(20)~~$ , and therefore :
$3$ is primitive root modulo $F_n(20)$ .
But , this proof strategy doesn't work in case when we want to find out if a $17$ is primitive root modulo $F_n(6)$ .