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Suppose that $f(x)$ is an irreducible cubic polynomial with integral coefficients. Suppose further that for all primes $p$, there exists an integer $n_p$ for which $p^2 \nmid f(n_p)$. Then it is a theorem of Erdős that $f(x)$ takes on infinitely many square-free values. This result was refined by Hooley, and extended to the situation where $f(p)$ takes on infinitely many square-free numbers where $p$ ranges over the rational primes by Helfgott in 2013, which was essentially proved in a different way by Reuss in 2015.

Consider, for a positive parameter $B$, the set $$\displaystyle S_f(B) = \{t \in \mathbb{N} : t \leq B, t \text{ is square-free}, \exists x \in \mathbb{Z} \text{ s.t. } f(x) = t\}.$$

I want to know the relative density of $S_f(B)$ in the set

$$\displaystyle T_f(B) = \{t \in \mathbb{N}: t \leq B, \exists x \in \mathbb{Z} \text{ s.t. } f(x) = t\},$$

that is, the quantity

$$\displaystyle \lim_{B \rightarrow \infty} \frac{\# S_f(B)}{\# T_f(B)}$$

if the limit exists.

It seems that the relative density should be higher than $6/\pi^2$, since there is a bias towards certain primes. In particular, if $t$ is a number which is divisible by a prime $p$ for which $f(x) \equiv 0 \pmod{p}$ has no solution, then $t$ cannot be in either set. Therefore, I suspect Chebotarev's density theorem to enter the picture somehow.

Any insight would be appreciated.

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    $\begingroup$ Why don't you expect the answer to be \prod_p (1 - (# of zeroes of f mod p^2)/p^2)? This converges because by Hensel there are O(1) solutions unless p | disc(f). Also away from these primes you can replace the zero count mod p^2 with it mod p instead. Now I'm confused because this is something 1/Dedekind zeta_K(2), except where you only take the product over degree 1 primes (here K = \Q[x]/f). So I bet this is wrong somewhere but I'll comment it anyway in case it helps! $\endgroup$ – alpoge Oct 7 '15 at 21:19

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