TITLE: Distribution of zeros of a polynomial mod. a prime
QUESTION [6 upvotes]: Let $\mathbb P$ be the set of prime numbers. 
Is there a non constant polynomial $f \in \mathbb Z[X]$ such that the set $$I_f := \{ \textstyle\frac{z}{p} : z \in \mathbb Z, p \in \mathbb P, p \mid f(z) \}$$
is dense in $\mathbb R$?
(I suppose that the following much stronger statement holds: Every polynomial $f \in \mathbb Z[X]$ which has a irreducible factor of degree at least $2$ satisfies the above condition.)

REPLY [2 votes]: In my paper, Polynomial congruences and density, Math Mag 80 (2007) 299-302, I prove this related result: 
Let $f(t)=t^eg(t)$ where $e$ is a nonnegative integer, $g$ is a polynomial of degree at least 2 with integer coefficients, and $g(0)\ne0$. Define $T_f$ by $$T_f=\lbrace r/m:\gcd(r,m)=1,{\rm\ and\ }m{\rm\ divides\ }f(r)\rbrace.$$ Then $T_f$ is dense in the reals. 
The proof uses nothing you wouldn't see in an undergrad intro Number Theory course (but of course it's weaker than what you want since I need all denominators, not just primes).