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While thinking about an unrelated problem, I had to decide whether $6$ was a primitive root with respect to multiple prime moduli. I could discover no obvious pattern as to primes for which $6$ is a primitive root: $11,13,17,41,59,61,\dots$; and those for which it is not: $19,23,29,31,37,43,47,53,67,71,73,\dots$. (The primes $2,3,5,7$ are not included in the analysis for the essentially trivial reason that $6\equiv 0,\pm 1$ with respect to those moduli.) After a little deeper look, I observed that $6$ is not a primitive root of prime $p\ge 11$ when $\exists n<p-1$ such that $2^n$ and $3^n$ are multiplicative inverses of each other $\bmod p$. When this occurred, it often, but not exclusively, occurred at $n=\frac{p-1}{2}$. For modest size primes, calculation is not an insurmountable obstacle, but as the primes grow in magnitude, calculation becomes correspondingly more tedious.

My question is this: Is there an easy way determine (short of actual calculation) for a given prime $p$ whether $6$ will be a primitive root?

Keith Backman
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    Well, you could test to see if $6$ is a quadratic residue. Primitive roots can't be quadratic residues. Beyond that...nobody knows how to determine the primes for which a given non-square is a primitive root. – lulu Aug 21 '21 at 16:05
  • I understand that the general problem may still be unapproachable, but I was hoping that since the primes I am interested in have the form $6m\pm 1$, there might be some obscure knowledge out there about the question as it applies to $6$ – Keith Backman Aug 21 '21 at 16:17
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    There is no natural number which is known to be a primitive root for infinitely many primes. I think you can do this if you assume GRH, but I doubt that comes with a useful construction. Maybe some density results. – lulu Aug 21 '21 at 16:24
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    See also http://oeis.org/A167794 – lhf Aug 21 '21 at 16:28
  • I think, you can do no better than to determine the order modulo $6$. If a number $N$ is given from which we do not know whether it is prime, we have to check the primality anyway. And the work is still not done. If we cannot factor $N-1$ , we might be unable to determine the order. – Peter Aug 21 '21 at 16:34
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    @KeithBackman You said you're interested in the primes that have the form $6m\pm 1$, but with the exception of $2$ and $3$, all primes have that form. – jjagmath Aug 21 '21 at 16:45
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    @jjagmath I was aiming for the commonality of the $6$ in the form of the primes and the $6$ for which I wish to know whether it is a primitive root. Think of it as if I were asking whether $a$ could be determined to be a primitive root with respect to primes of the form $am\pm 1$, and then focusing on $a=6$. – Keith Backman Aug 21 '21 at 17:20
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    Your observation about $2^n$ and $3^n$ is nearly trivial. They multiply to $6^n$ and the order of a base is the LCM of the order of each of it's prime factors at greatest. – Roddy MacPhee Aug 21 '21 at 17:35
  • Course we can note that $6^{-1}\equiv 5m+1\pmod {6m+1}$ and $6^{-1}\equiv m\pmod {6m-1}$ and use properties of inverses to infer things. – Roddy MacPhee Aug 21 '21 at 18:07
  • Additive inverse only have same order when $-1$ shows up in an even position, multiplicative inverses only have same order If I recall correctly. – Roddy MacPhee Aug 21 '21 at 19:15
  • $6^{3m-1}\equiv m\pmod{6m+1}$ and $6^{6m-3}\equiv m \pmod{6m-1}$ interesting. – Roddy MacPhee Aug 21 '21 at 19:48
  • @RoddyMacPhee $6^{3m-1} \not\equiv m \pmod {6m+1}$ for $m=3$ – jjagmath Aug 21 '21 at 20:06
  • Okay maybe I screwed up I was using the facts that $m$ is either the multiplicative inverse or it's negation for $6$ so ascribing discrete logs ... But your observation is equivalent to $19$ not working ... – Roddy MacPhee Aug 21 '21 at 20:13
  • It depends on what you mean by calculation. I consider any kind of a test a calculation so my answer would be no. One way to determine it is to try for all $d|(p-1)$ whether $6^d=1\ (mod\ p)$, and $6$ is a primitive root $mod\ p$ if-and-only-if $d=p-1$ is the only value of $d$ that satisfies this condition. – Steven Clark Aug 21 '21 at 20:46
  • @Steven that's the definition. – Roddy MacPhee Aug 22 '21 at 00:51
  • I believe the definition of a primitive root $mod\ p$ where $p\in\mathbb{P}$ is the powers of a primitive root generate all of the elements of the multiplicative group $1..p-1$. I'm not sure if there's a more efficient test. The question here is related to the recent and more general question at https://math.stackexchange.com/questions/124408/finding-a-primitive-root-of-a-prime-number. – Steven Clark Aug 22 '21 at 02:00

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