A question on how polynomials split over $\mathbb{F}_p$ for large primes $p$

Question

Suppose that $f(x) \in \mathbb{Z}[x]$ is an irreducible polynomial (over $\mathbb{Q}$). Let $p$ be a very large prime with respect to the coefficients of $f$. Then it is possible that $f(x)$ may factor over $\mathbb{F}_p$, and indeed $f$ may split completely over $\mathbb{F}_p$. My question concerns whether $f$ may have 'small roots' over $\mathbb{F}_p$. That is, we say a root $r$ of $f$ over $\mathbb{F}_p$ (if one exists) is 'small' if (by abuse of notation) the smallest positive integer representation $r \ll p^{1/d}$, where $d = \deg(f)$. Can we characterize, for a given $f$, the set of primes such that $f$ has a small root over $\mathbb{F}_p$?

Answer 1

If $r \ll p^{1/d}$ and $p \mid f(r)$ then (since $f(r) \neq 0$) $f(r) = ap$ for some nonzero $a \ll 1$. Hence for each of finitely many choices of $a$ we are asking for prime values of $f(r)/a$ as $r$ ranges over ${\bf Z}$. That's a characterization of sorts, though (for each choice of implicit constant $C$ in $|r| \leq C p^{1/d}$)
the question of whether there are infinitely many such $r$ comes down to the Bunyakovsky Conjecture, which still seems well out of reach.