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$2$ 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 $m$.

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  • $\begingroup$ If $p\equiv 1\pmod{8}$, then $2$ is a quadratic residue modulo $p$, hence not a primitive root. $\endgroup$ – pisco Dec 14 '17 at 14:20
  • $\begingroup$ @pisco125, Could you please elaborate on that? $\endgroup$ – Elimination Dec 14 '17 at 14:22
  • $\begingroup$ @BarryCipra, you're right - $n\ge 2$ (I'll add that) $\endgroup$ – Elimination Dec 14 '17 at 14:32
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$$2^{2^n}\equiv -1\pmod {F_n}$$ Then $$2^{2^{n+1}}\equiv 1\pmod{F_n}$$ And clearly $2^{n+1}<F_n-1$ for $n\ge 2$

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  • $\begingroup$ Oh wow, nice solution! Thank you! $\endgroup$ – Elimination Dec 14 '17 at 14:25

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