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Is this part of a more general claim? I have dug through the textbook results and couldn't find one that shows this. i.e., that given a primitive root $a$, $a^n$ is a primitive root iff $\gcd(n,p-1)=1$.

^^I don't get why we need the gcd part as $$a^{n(p-1)}\equiv 1 \pmod p$$ seems to be the case for any $n$, so I don't get where the coprimality comes in.

Does it have anything to do with the fact that if $n$ were coprime with $p-1$, then $n$ would be in it's reduced residue system?

Jason Xu
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    Write down the definition of a primitive root then check it carefully here, that's all. – Qiaochu Yuan Oct 24 '23 at 06:00
  • Pardon me for interjecting, but I meant to ask this in a non-group theoretic context, i.e., prove only with number theory methods – Jason Xu Oct 24 '23 at 06:50
  • What do you mean by “number theory methods”? – MJD Oct 24 '23 at 07:48
  • You are making it too difficult for yourself, if you want to use $p=139$. As is often the case, a smaller number is already illuminating. Take $p=7$. Show first that $3$ is a primitive root (follow Qiaochu's advice from above). Then check that $3^n$ is not a primitive root when $n=2,3$ or $4$, but $3^5\equiv5$ again is a primitive root. Notice that $2,3,4$ share a prime factor with $p-1=6$. The real exercise is to figure out how that relates to the non-primitivity?? – Jyrki Lahtonen Oct 24 '23 at 08:16

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