$2$ cannot be a primitive root of a prime $F_n$

Question

$2$ cannot be a primitive root of a prime $F_n = 2^{2^n} + 1$ where $n\ge 2$

I've understood that the fact that $F_n \equiv 1 \pmod{8}$ for $n\ge 2$ might be helpful here, but I don't see how (Though I'm open to other suggestions of course)

I have seen in class that there's a primitive root modulo $m$ iff:

$m=2,4$
$m=p^k$, $p$ an odd prime and $k\ge 1$
$m=2p^k$, $p$ an odd prime and $k\ge 1$

In particular, for $m=8$ there's no primitive root modulo $m$.

If $p\equiv 1\pmod{8}$, then $2$ is a quadratic residue modulo $p$, hence not a primitive root. — pisco, Dec 14 '17 at 14:20

score 2 · Accepted Answer · answered Dec 14 '17 at 14:24

2

$$2^{2^n}\equiv -1\pmod {F_n}$$ Then $$2^{2^{n+1}}\equiv 1\pmod{F_n}$$ And clearly $2^{n+1}<F_n-1$ for $n\ge 2$

answered Dec 14 '17 at 14:24

ajotatxe

65,084

Oh wow, nice solution! Thank you! – Elimination Dec 14 '17 at 14:25

$2$ cannot be a primitive root of a prime $F_n$

1 Answers1

Linked