Sign up ×

Browning in this paper proves that if $f \in \mathbb{Z}[x]$ is an irreducible polynomial of degree $d \geq 3$ and $k$ is an integer $\geq 3d/4 + 1/4$, then we have $$\#\{n \in \mathbb{Z} \cap [1, X] : f(n) \text{ is $k$-free}\}\sim C_{1}(k, f)X$$ as $X \rightarrow \infty$ where $$C_{1}(k, f) = \prod_{p}\left(1 - \frac{\varrho(p^{k})}{p^{k}}\right)$$ and $\varrho_{f}(n) = \#\{a \pmod{n}: f(a) \equiv 0 \pmod{n}\}$. Is there anywhere that gives a similar asymptotic for the case when $f$ is reducible?

share|cite|improve this question
The same asymptotic is conjectured to hold for arbitrary $f$, reducible or not. Following an idea of Granville, it is possible to prove that it follows from the ABC conjecture. (See section 12.2 of "Heights in diophantine geometry" by Bombieri and Gubler). However, even the existence of infinitely many square-free values of $n^4 + 1$ is not known unconditionally. On the other hand, using an elementary sieve argument, it is possible to prove the conjecture for the case when $f$ splits into factors of degree at most 2. (And, I believe, even of degree $\leq 3$, although this is more difficult.) – Vesselin Dimitrov Feb 2 '13 at 14:40
@lzk712 tried to fix the tex by escaping backslashes. – joro Feb 2 '13 at 15:36
What is $d$? The degree of $f$? – Gerry Myerson Feb 2 '13 at 23:01
@Gerry Myerson: Yes $d$ is the degree of $f$. Thanks!; @joro: Thanks for the formatting fix.; @Vesselin Dimitrov: Do you know where one can find the case when $f$ splits into factors of degree at most 2 worked out? Everyone seems to just say by "sieve methods". – lzk712 Feb 3 '13 at 1:28
Take a look at the posting of Ravi B on this link:… . For k-frees, the argument works assuming $f$ splits into factors of degree at most k. As for the Browning result you quote, my guess is that the proof would likewise work assuming $f$ splits into irreducibles whose degrees satisfy the mentioned inequality. – Vesselin Dimitrov Feb 3 '13 at 1:52

1 Answer 1

The reducible case is not much harder than the irreducible case, if one uses an argument in Greaves's paper (used originally by Gouvea and Mazur) "Power-free values of binary forms". In particular, the main term can be defined just as easily for reducible polynomials $f(x) \in \mathbb{Z}[x]$. The error terms can be dealt with simply as follows. Write $f(x) = f_1(x) \cdots f_s(x)$, where $f_i(x)$ is irreducible over $\mathbb{Z}$ and define the error term $E_i(X) = \# \{x \in \mathbb{Z} : p^k | f(x) \text{ for some } x > \xi\}$, where $\xi = \frac{1}{k} \log X$. Then the overall error term is just $E(X) = \sum_{j=1}^s E_i(X)$. Thus one can use whatever techniques one would like (in this case, Browning's result is the strongest; although using a different determinant method I am able to reproduce the same result) to estimate each of the $E_i(X)$ and multiply the result by $s$, which is negligible since it is an absolute constant and the error term will be a power saving over the main term (assuming the constant in front of the main term is non-zero).

I should remark that the above estimates do not cover the case when say $f_1(x) = uv^{k-l}$ and $f_2(x) = u'v^l$, or when three factors of $f$ conspire to give us a non-$k$-free term. However, this situation is easily shown to give a negligible contribution. To see this, consider the equations $f_1(x) = uv^{k-l}$, $f_2(x) = u'v^l$ and $f(x) = wv^k$. With $x$ and $v$ fixed, we see that $u, u'$ are divisors of $w$ and hence there are no more than $d(w) = O(w^\epsilon)$ many choices. A simple analysis on the size of $w$ yields that $w = O(X^d)$, and so $d(w) = O(X^\epsilon)$. We may then crudely estimate the number of points lying on the intersection of the varieties defined by $f_1(x) = uv^{k-l}$ and $f_2(x) = u'v^l$ where we take $u, u'$ to be fixed as follows. For $u, u'$ fixed, each of the equations define a curve in $\mathbb{A}^2$, and they have at most finitely many intersection points. Thus the overall contribution is $O(X^\epsilon)$ and is negligible.

share|cite|improve this answer
It's also worth noting that only finitely many primes (those dividing the discriminant of $f$) can simultaneously divide two of the $f_j(n)$. In fact, if one partitions the domain into suitable arithmetic progressions, one can avoid that possibility altogether - see Lemma 3.2 of – Greg Martin Sep 19 '14 at 1:49

Your Answer


