We know that even squares cannot be primitive roots modulo primes.Are all other natural numbers primitive roots mod some p?

My heuristic argument goes as follows: the probability that a natural number is a primitive root mod p is $\frac{\varphi(p-1)}{p}$.Therefore, the probability that a number is not a primitive root mod any prime p<=x is:

$$Prob_x=\prod_{p\leq x}{1-\frac{\varphi(p-1)}{p}} $$

which is less than

$$\zeta_x(1)^{-1}=\prod_{p\leq x}{1-\frac{1}{p}} $$

So as x goes to infinity, $Prob_x$ will go to $0$, but that only proves (I think) that the density of the non-primitive roots is 0, not that they are only the even squares.

EDIT I did some research on Artin's conjecture on primitive roots and I found that a number $n$ is a primitive root mod p iff p does not split in any of the fields $L_q=Q(\zeta_q,n^\frac{1}{q})$, where q is a prime dividing p-1.I guess if we apply Chebotarev we can find a prime for which n is a primitive root (in fact using Chebotarev we can estimate the expected density of primes such that n (non-square) is a primitive root, namely $$C_{Artin}=\prod (1-\frac{1}{q.(q-1)})=0.37395...$$

I am not exactly sure why n would be a primitive root mod p iff p splits completely in those fields (I guess Dedekind's theorem?; and something about $n^\frac{p-1}{q}$ not being 1 mod p).

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    $\begingroup$ Look up Artin's conjecture for related information. $\endgroup$ – Jyrki Lahtonen Jan 12 '17 at 22:36
  • $\begingroup$ Wow, that is a lot stronger than my question.Is there way to prove it without resorting to that (especially since it is still unproven)? $\endgroup$ – Bogdan Simeonov Jan 12 '17 at 22:39

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