If $q\in\mathbb{Z}^+$ is not a perfect square, does there always exist an odd prime $p$ such that $q$ is a generator of $\mathbb{Z}/p\mathbb{Z}^\times$? Can we find always find infinitely many such $p$?
Are all non-squares generators modulo some prime $p$?
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elementary-number-theory
1 Answers
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This is a conjecture of Artin. Hooley proved the conjecture is true for all nonsquare $q > 1$ if the generalized Riemann hypothesis is true for zeta-functions of number fields. Later work by Hooley, following work by R. Murty and Gupta, established the conjecture unconditionally (i.e., with no GRH assumptions) for all prime $q$ with at most 2 exceptions (and nobody expects there really are any exceptions at all). For example, the conjecture is provably true for at least one of the choices $q = 2$, 3, or 5 (and surely is true for all three!) but we can't pin down a specific one of those three values for which the conjecture is definitely true. To this day the conjecture is not proved for any specific value of $q$.
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0In what sense it is *now* almost proved? The wikipedia page does not list any recent developments. – 2011-05-16
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0Heath-Brown's proof is certainly a lot more recent than the conjecture itself. Didn't mean to imply any breaking news from last week. – 2011-05-16
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0Right, thanks . – 2011-05-16
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0Alon: I edited your answer. The conjecture is proved for all but at most two *prime* values of $q$, not for all but at most two values of $q$ in general. Look at the wikipedia page again when it discusses that point. – 2011-05-16
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0@Alon and others: you've discussed the second of Thomas' questions, but what about the first? Is it known that for every positive non-square $q$ there's a prime $p$ for which $q$ is a primitive root? – 2011-05-17
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0@Gerry, that's a good point. I thought about this a little and didn't get much further than observing that you want to demand $p>2$ lest the problem become trivial. – 2011-05-18