# square-free parts of values of polynomials

Given a polynomial $f(x) \in \mathbb{Z}[x]$ of degree $d$, consider the following three sets:

$$N_1(x) = \#\{k \leq x: f(k) \text{ is square-free}\}$$ $$N_2(x) = \#\{n \leq x: n = f(k) \text{ is square-free}\}$$ $$N_3(x) = \#\{n \leq x: n \text{ is the square-free part of f(k)}\}$$

Generally, we should have that $N_1(x) \sim c x$ (not proven for large degrees, assumes that $f$ satisfies the necessary congruence conditions). Then $N_2(x) \sim c' x^{1/d}$. What do we know about the growth of $N_3(x)$? Do we get significantly more values by considering square-free parts and not just square-free values?

Any references for this would be appreciated.

Edit: Changed from big-O notation to make order of growth more precise.

• In the third part, do you consider $k\leq x$ or all $k$? Mar 4, 2014 at 4:06
• For all $\epsilon$, for almost all $x$, the squarefree part of $x$ is greater than $x^{1-\epsilon}$. I think this implies that you do not get an exponential improvement, at least as the only reasonable conjecture. Mar 4, 2014 at 4:14
• Thanks Will! Yes, for the third, I'm considering all $k$. What you wrote makes a lot of sense... there aren't many integers divisible by large squares, and so allowing square-free parts shouldn't help all that much.
– stl
Mar 4, 2014 at 5:13
• You probably should write $N_1(x) = \Omega(x)$ or $N_1(x) \gg x$, and similarly for $N_2(x)$; the notation $O(x)$ indicates only an upper bound. Mar 4, 2014 at 7:19
• @bean No, but it's a quick argument using either sieves or Euler products, I think. Oct 3, 2020 at 23:50