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It is known that every prime $p$ has a primitive root modulo $p$. Is every number $a$ which is not a perfect square a primitive root modulo $p$ for some prime $p$? If it is a square, we already have $a^{\frac{p - 1}{2}} \equiv 1 \text{ (mod $p$)}$, so the condition on not being a square is necessary. However, I'm not sure what would suggest that not being a perfect square is sufficient.

modnar
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    Yes, but it isn't easy to show. The general problem (for nth powers rather than just squares) is discussed here: [http://math.stackexchange.com/questions/6976/proving-that-an-integer-is-the-n-th-power]. – FredH Jun 16 '13 at 20:35

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I believe that the full answer is not yet known. Certainly the (much) more difficult problem of whether every number other than $-1$ and the perfect squares is a primitive root of infinitely many primes is not fully settled. This is Artin's Conjecture.

However, there are many partial results towards Artin's Conjecture. for example, Hooley proved Artin's Conjecture, assuming an (unproved) version of the Generalized Riemann Hypothesis.

André Nicolas
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