If 2 is a primitive root of $p$, is it also a primitive root of $p^2$? If $2$ is a primitive root of an odd prime $p$, is it also a primitive root of $p^2$?
Edit: For the first 10,000 primes $p$, Wolfram|Alpha confirmed that $2$ is a primitive root of $p$ if and only if $2$ is also a primitive root of $p^2$.
 A: Note: This answer is not complete. It may not even be correct. But it has generated too many comments for me to remove it, and it may at least be of interest as an avenue of proof that doesn't work (unless someone can fix it). So I am leaving it up.
Clearly, since $2^k\equiv 1 \bmod p^2$ implies $2^k\equiv 1\bmod p$, the period of $2$ modulo $p^2$ must be a multiple of the period modulo $p$, i.e., $p-1$. On the other hand, the period must divide $\phi(p^2)=p(p-1)$, so it must equal $p(p-1)$ unless it equals $p-1$, in which case $p^2\mid 2^{p-1}-1$.
But if it is, then $p^2\mid (2^{(p-1)/2}-1)(2^{(p-1)/2}+1)$. $p$ does not divide the first factor by assumption, so $2^{(p-1)/2}\equiv-1\bmod p^2$ … and the conclusion is still not clear.
A: Hint:
$$\text{if $\,2\,$ is a primitive root modulo $\,p\,$ and }\;\;2^{p-1}\neq 1\pmod{p^2}$$
$$\,\,\text{then $\,2\,$ is a primitive root modulo}\,\,p^2$$
In case $\,2^{p-1}=1\pmod{p^2}\,$ no problem: $\,2+p\,$ is a primitive root modulo $\,p^2\,$
Spoiler!

 Read the first part of this paper

A: It's expected (I think) that there are infinitely many counterexamples, but that's not known. In general, there's no reason to expect that a given primitive root modulo $p$ should always be a primitive root modulo $p^2$.
(Originally I claimed that we actually know two counterexamples, 1093 and 3511; but that's wrong - 2 is a cube modulo 1093 and a square modulo 3511, hence not a primitive root for either prime.) 
