What's the probability that k + n^2 is squarefree, for fixed k?

2010-03-01T22:25:36Z

While playing around with this question (when is the sum of two squares squarefree?), from some experimental computations (and bolstered by the fact that the density of squarefree positive integers is known to exist), I came up with the following conjecture: the asymptotic density of squarefree numbers in the sequence $(k+1^2, k+2^2, k+3^2, \ldots)$, for fixed k, exists and depends on k.

To give an example of what I mean, consider numbers of the form $1 + n^2$. 895 of the numbers $1+1^2, 1+2^2, \ldots, 1+1000^2$ are squarefree; 897 of the next thousand are; 895 of the third thousand; 896 of the fourth thousand. 891 of the numbers $1+1000001^2, \ldots, 1+1001000^2$ are squarefree, as are 895 of the numbers $1+2000001^2, \ldots, 1+2001000^2$. So there seems to be some constant $C_1$, probably between 0.89 and 0.90, such that $1+k^2$ has ``probability'' $C_1$ of being squarefree. The same thing seems to happen if we replace 1 with other integers, although with other constants.

Are such constants known to exist? If they are, how can they be computed?

Answer by Bjorn Poonen for What's the probability that k + n^2 is squarefree, for fixed k?

2010-03-01T22:48:02Z

More generally, suppose that $f(x) \in \mathbf{Z}[x]$ has no repeated factors. For each prime $p$, let $c_p$ be the number of integers $x \in \{0,1,\ldots,p^2-1\}$ satisfying $f(x) \equiv 0 \pmod{p^2}$. Heuristically, the probability that a random integer $x$ is such that $f(x)$ is not divisible by $p^2$ equals $1-c_p p^{-2}$, and these conditions should be independent by the Chinese remainder theorem, so one would conjecture that the fraction of integers $x$ in $\{1,2,\ldots,N\}$ such that $f(x)$ is squarefree should tend to $\prod_p (1-c_p p^{-2})$, where the product is taken over all primes $p$. For large $p$, we have $c_p \le \deg f$, so this product converges.

This guess has been proved for $\deg f \le 3$ (the case $\deg f=3$ is a nontrivial result of C. Hooley). In particular, this answers your question for $f(x)=x^2+k$. There is no $f$ of degree $4$ or greater for which the density is known to exist (except in cases when the density is $0$ because some $c_p$ equals $p^2$). On the other hand, A. Granville proved that the $abc$ conjecture implies that the density exists and equals the predicted value for $f$ of any degree. For further references and a generalization to multivariable polynomials, see the papers cited in the references below.

C. Hooley, On the power free values of polynomials, Mathematika 14 (1967), 21-26.

A. Granville, $ABC$ allows us to count squarefrees, Internat. Math. Res. Notices 1998, no. 19, 991-1009.

B. Poonen, Squarefree values of multivariable polynomials, Duke Math. J. 118 (2003), no. 2, 353-373.

What's the probability that k + n^2 is squarefree, for fixed k? - MathOverflow

What's the probability that k + n^2 is squarefree, for fixed k?

Answer by Bjorn Poonen for What's the probability that k + n^2 is squarefree, for fixed k?