There is a simple answer here, so someone might as well record it.

Let $n$ be a nonzero integer. If $n = -1$ or $n$ is a square then there is no odd prime number $p$ such that $n$ is a primitive root modulo $p$. There are no other obvious obstructions. (It is worth thinking for a second why we do not have to rule out $n$ being a cube, for instance: this is a nice exercise in cyclic group theory.)

There is a famous conjecture that these obvious necessary conditions are the only ones: namely Artin's Primitive Root Conjecture asserts that for any integer $n$ which is not $0$, $-1$ and not a square, there are infinitely many prime numbers $p$ such that $n$ is a primitive root modulo $p$. In fact the conjecture is more precise than this: the set of primes $p$ for which such an $n$ is a primitive root is conjectured to have positive relative density among all primes and, at least under some mild additional restrictions, this density is conjectured to be a certain specific number which is independent of $n$:

$ C = \prod_{p \text{ prime}} \left(1- \frac{1}{p(p-1)} \right)$;

this $C$ is known as Artin's constant. This conjecture was proved by C. Hooley in 1967 assuming the Generalized Riemann Hypothesis. More recently unconditional results have been given by Gupta, R. Murty and Heath-Brown which consider several numbers $n$ at a time and show that Artin's Conjecture must be true for at least one of them. But the conjecture is still open for any one fixed value of $n$.

There is a simple answer here, so someone might as well record it.

Let $n$ be a nonzero integer. If $n = -1$ or $n$ is a square then there is no prime $p > 3$ such that $n$ is a primitive root modulo $p$. There are no other obvious obstructions. (It is worth thinking for a second why we do not have to rule out $n$ being a cube, for instance: this is a nice exercise in cyclic group theory.)

There is a famous conjecture that these obvious necessary conditions are the only ones: namely Artin's Primitive Root Conjecture asserts that for any integer $n$ which is not $0$, $-1$ and not a square, there are infinitely many prime numbers $p$ such that $n$ is a primitive root modulo $p$. In fact the conjecture is more precise than this: the set of primes $p$ for which such an $n$ is a primitive root is conjectured to have positive relative density among all primes and, at least under some mild additional restrictions, this density is conjectured to be a certain specific number which is independent of $n$:

$ C = \prod_{p \text{ prime}} \left(1- \frac{1}{p(p-1)} \right)$;

this $C$ is known as Artin's constant. This conjecture was proved by C. Hooley in 1967 assuming the Generalized Riemann Hypothesis. More recently unconditional results have been given by Gupta, R. Murty and Heath-Brown which consider several numbers $n$ at a time and show that Artin's Conjecture must be true for at least one of them. But the conjecture is still open for any one fixed value of $n$.

There is a simple answer here, so someone might as well record it.

Let $n$ be a nonzero integer. If $n = -1$ or $n$ is a square then there is no odd prime number $p$ such that $n$ is a primitive root modulo $p$. There are no other obvious obstructions. (It is worth thinking for a second why we do not have to rule out $n$ being a cube, for instance: this is a nice exercise in cyclic group theory.)

There is a famous conjecture that these obvious necessary conditions are the only ones: namely Artin's Primitive Root Conjecture asserts that for any integer $n$ which is not $0$, $-1$ and not a square, there are infinitely many prime numbers $p$ such that $n$ is a primitive root modulo $p$. In fact the conjecture is more precise than this: the set of primes $p$ for which such an $n$ is a primitive root is conjectured to have positive relative density among all primes and, at least under some mild additional restrictions, this density is conjectured to be a certain specific number which is independent of $n$:

$ C = \prod_{p \text{ prime}} \left(1- \frac{1}{p(p-1)} \right)$;

this $C$ is known as Artin's constant. This conjecture was proved by C. Hooley in 1967 assuming the Generalized Riemann Hypothesis. More recently unconditional results have been given by Gupta, R. Murty and Heath-Brown which consider several numbers $a$ at a time and show that Artin's Conjecture must be true for at least one of them. But the conjecture is still open for any one fixed value of $a$.

There is a simple answer here, so someone might as well record it.

Let $n$ be a nonzero integer. If $n = -1$ or $n$ is a square then there is no odd prime number $p$ such that $n$ is a primitive root modulo $p$. There are no other obvious obstructions. (It is worth thinking for a second why we do not have to rule out $n$ being a cube, for instance: this is a nice exercise in cyclic group theory.)

There is a famous conjecture that these obvious necessary conditions are the only ones: namely Artin's Primitive Root Conjecture asserts that for any integer $n$ which is not $0$, $-1$ and not a square, there are infinitely many prime numbers $p$ such that $n$ is a primitive root modulo $p$. In fact the conjecture is more precise than this: the set of primes $p$ for which such an $n$ is a primitive root is conjectured to have positive relative density among all primes and, at least under some mild additional restrictions, this density is conjectured to be a certain specific number which is independent of $n$:

$ C = \prod_{p \text{ prime}} \left(1- \frac{1}{p(p-1)} \right)$;

this $C$ is known as Artin's constant. This conjecture was proved by C. Hooley in 1967 assuming the Generalized Riemann Hypothesis. More recently unconditional results have been given by Gupta, R. Murty and Heath-Brown which consider several numbers $n$ at a time and show that Artin's Conjecture must be true for at least one of them. But the conjecture is still open for any one fixed value of $n$.