Let $n>1$ be an non square positive integer (you can have it prime, if you wish), does there exist a prime $p>2$ such that $n$ generates the multiplicative group of $\mathbb F_p$? It sounds true, but I could not find an immediate proof for that... maybe using some reciprocity law? Not sure.
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Have you tried a few examples? Does it work when p=3? – User0112358 Dec 15 '15 at 21:58
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If n is odd let p=2. Not sure about even n. – DanielWainfleet Dec 15 '15 at 21:59
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p=3, n=7 fails since 7 is congruent to 1 modulo 3 so it does not generate. But anyway the question is the other way round: you should fix n and find p – Reyx_0 Dec 15 '15 at 22:03
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edited with p>2, sorry. – Reyx_0 Dec 15 '15 at 22:04
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1You may find this useful as a start. Every prime $\gt 2$ does have a prime primitive root. There is a lot more known, including size estimates. – André Nicolas Dec 15 '15 at 22:23
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Related: https://math.stackexchange.com/questions/422201 – Watson Nov 23 '18 at 19:50
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The general question in a strong form is the contents of Artin's Conjecture:
Every integer which is neither a perfect square nor equal to $−1$ is a primitive root modulo infinitely many primes.
This remains unproved.
lhf
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