Squareful values of polynomials

Question

Recall that an integer $n$ is called squareful if for every prime $p$ with $p \mid n$, we also have $p^2 \mid n$.

Any squareful number can be written uniquely as $n= x^2 y^3$ where $y$ is squarefree. From this, it is easy to see that $$\#\{ n \in \mathbb{Z}: |n| \leq X, \, n \text{ is squareful} \} \ll X^{1/2}.$$

I would like a version of this for polynomials.

Let $f \in \mathbb{Z}[x]$ be non-constant and separable. Then does there exist $\delta > 0$ such that $$\#\{ n \in \mathbb{Z}: |n| \leq X, \, f(n) \text{ is squareful} \} \ll X^{1 - \delta} \quad ?$$

Hopefully there are not some necessary local conditions here that I overlooked. In my application I am happy to change $f$ as required so that one can assume that $f$ is sufficiently "general". Moreover, I can even assume that $f$ is of very large degree if necessary to simplify things.

I would normally try to prove something like this using the large sieve, however the large sieve gives the poor upper bound $X/(\log X)$, whereas I would like a power saving.

If necessary, I'm happy to assume some standard conjectures (e.g. the abc conjecture).

Answer 1

If $f$ is squarefree and of degree at least three, then the abc conjecture implies $f(x)$ is squareful finite number of times. For reference and several papers of Granville:

Power free-values of polynomials. Langevin noted in [Lan2] the following conjecture which is a consequence of the abc conjecture. Let $F(X)$ be a polynomial with integer coefficients and no repeated roots. For any $\epsilon > 0$, there exists a constant $C_{\epsilon,F}$ such that for any integer $n$,$|n|^{\deg(F)-1-\epsilon} < C_{\epsilon,F} rad(F(n))$.

Actually abc implies the squareful part of $f(x)$ can't be too large infinitely often. If the squareful part of $f(x)$ is $s$, then $ s < C_\epsilon x^{2+\epsilon}$.

Answer 2

In the sieve book of Cojocaru and Murty they give a simple application of the square sieve of Heath-Brown, namely in Theorem 2.3.5 of their book they prove that $$\#\{1\leq n \leq x:f(n)=\square\}\ll_{f,\epsilon} x^{1-\epsilon}$$ for all $0<\epsilon<\frac{1}{3}$. Squares and squarefulls have the same asymptotic density and there is definitely a chance to make the square-sieve approach work for $$\#\{1\leq n \leq x:f(n)=\text{ squarefull}\}$$ without abc. Note that if $f(n)$ is square-full then $f(n)=a^2b^3$ hence there exists $b \in \mathbb{N}$ such that $f(n)b$ is a square. The case $b=1$ is what they do however their proof can be easily adopted to work for $bf(n)$. It gives a bound independent of $b$ by changing $\mathcal{P}$ in their proof to be the set of all primes in $(x^{1/3},2x^{1/3}]\setminus \{p \text{ prime } : p|b\}$. The prime divisors of $b$ that are excluded are at most $\log b\ll \log x$, while the set of primes is $\asymp x^{1/3} (\log x)^{-1}$, therefore all estimates go through unaltered. This does not solve your problem since one has to do something extra to get more saving when $b$ gets larger, but there is some hope.