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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.)

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The set $I_f$ is dense for quadratic polynomials which are irreducible over $\mathbb{Q}$, by the work of Duke-Friedlander-Iwaniec (Ann. of Math. 141 (1995), 423-441) and Toth (IMRN 2000, No. 14, 719-739). In fact we have uniform distribution even when $p$ is restricted to a reduced residue class of any modulus.

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  • $\begingroup$ Edit : I changed the notation $I_p$ into $I_f$ in order to reflect the change of notation in the question. $\endgroup$ – François Brunault Apr 11 '11 at 16:39
  • $\begingroup$ Thanks! Do you know why $\mathbb{Q}$ appeared differently in my original response? $\endgroup$ – GH from MO Apr 11 '11 at 16:51
  • $\begingroup$ @GH : I didn't notice that (and I haven't changed this part of the code). $\endgroup$ – François Brunault Apr 11 '11 at 18:23
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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).

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