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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 $F_n(6)$ .

  • 1
    How do you get from "3 is a quadratic nonresidue mod $F_n(20)$" to "3 is a primitive root modulo $F_n(20)$"?2012-02-02

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