Approximate primitive roots mod p Artin conjectured that if $a$ is an integer which is not a square and not $-1$ then $a$ is a primitive root for infinitely many primes.  This conjecture has not been resolved, but partial results are known:  Heath-Brown showed that there are at most two prime numbers $a$ for which the conjecture fails.
I'd like to know if a different kind of partial result is known.  Let $I(p)$ denote the index of the subgroup of $(\mathbf{Z}/p\mathbf{Z})^{\times}$ generated by 2.  Thus $I(p)=1$ if and only if 2 is a primitive root mod $p$.  Can one show that there is an infinite sequence of primes in which $I$ remains bounded?
 A: Here is something that is much weaker than what you are asking. The proof is elementary (but not entirely trivial). For every $\epsilon>0$, the series
$$ \sum_p \frac{I(p)^\epsilon}{p^{1+\epsilon}} $$
converges. For example, this implies that for every $N>0$, the set of primes $p$
satisfying
$$ I(p)>\frac{p}{(\log\log p)^N} $$
has (analytic) density zero.
A: A result of Erdos and Murty asserts that if $\epsilon(p)$ is any decreasing function tending to zero, then $I(p) \leq p^{1/2-\epsilon(p)}$ for almost all primes $p$ (i.e., all but $o(\pi(x))$ primes $p \leq x$). 
Kurlberg and Pomerance (see Lemma 20 in the paper mentioned below) show that for a positive proportion of primes $p$, one has the stronger bound $I(p) \leq p^{0.323}$. This follows from a result of Baker and Harman on shifted primes with large prime factors.
The Erdos--Murty paper is #77 at
http://www.mast.queensu.ca/~murty/index2.html
and the Kurlberg--Pomerance paper is
http://www.math.dartmouth.edu/~carlp/PDF/par13.pdf
See also Theorem 23 of this paper (which is conditional on GRH).
