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

$2$ 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:

1. $m=2,4$
2. $m=p^k$, $p$ an odd prime and $k\ge 1$
3. $m=2p^k$, $p$ 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
• @pisco125, Could you please elaborate on that? – Elimination Dec 14 '17 at 14:22
• @BarryCipra, you're right - $n\ge 2$ (I'll add that) – Elimination Dec 14 '17 at 14:32

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