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Artin's conjecture on primitive roots states that a given integer a that is neither a perfect square nor −1 is a primitive root modulo infinitely many primes p. Can we expect that it is also a primitive root modulo infinitely many squares of primes?

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Yes. While being a primitive root (mod $p$) does not guarantee being a primitive root (mod $p^2$), it makes it extremely likely. (The proportion of exceptions for a given prime $p$ is $1/p$.) Therefore we expect that for the vast majority of primes $p$ for which $a$ is a primitive root, $a$ is also a primitive root (mod $p^2$). Almost equivalently, we expect there to be very few primes $p$ such that $p^2\mid(a^p-a)$. However, all of these statements are beyond our current ability to prove, I believe.

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  • $\begingroup$ This seems like a perfectly fine answer for a perfectly fine question. I'm confused why it is closed. The weaker question as to whether $2^{p-1} \not\equiv 1 \mod p^2$ for infinitely many $p$ still remains open. $\endgroup$ – user760870 Jul 12 at 16:56

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