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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 $p$?

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up vote 11 down vote accepted

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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In what sense it is now almost proved? The wikipedia page does not list any recent developments. – lhf May 16 '11 at 16:28
Heath-Brown's proof is certainly a lot more recent than the conjecture itself. Didn't mean to imply any breaking news from last week. – Alon Amit May 16 '11 at 16:37
Right, thanks . – lhf May 16 '11 at 16:57
Alon: 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. – KCd May 16 '11 at 21:40
That's right - thanks! – Alon Amit May 16 '11 at 21:50

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