# Squarefree numbers $n$ such that $432n+1$ is also squarefree

This is a second attempt (see Primes $p$ such that $432 p +1$ is prime)

Is the set of squarefree numbers $n$ such that $n(432 n+1)$ is also squarefree known to be infinite?

Fact: the number of such numbers $n$ such that $n\leq 10^6$ is precisely $553095$. Do we expect that $$\lim_{x\to \infty}\frac{\# \{n \leq x\ : \ n \ \textrm{is squarefree integer, and } 432n+1 \ \textrm{is squarefree}\}}{x} = 1/2?$$

• I don't have time to work this out, but the density is the product of local densities, which is surely not $1/2$ but some more complicated number (probably transcendental). That is, for each prime $p$ let $c_p$ denote the density of $n$'s such that $n(432n+1)$ is not divisible by $p^2$, and then the density in the post will be $\prod_p c_p$. – GH from MO Aug 5 '14 at 13:51
• What is so special about 432? – Emil Jeřábek Aug 5 '14 at 13:52
• See my response here how to do this for $n(n+1)$: mathoverflow.net/questions/172009/… – GH from MO Aug 5 '14 at 13:54
• I think you should put more about the motivation in the question, then. – Emil Jeřábek Aug 5 '14 at 13:55
• Actually, thinking a bit about it, the problem is not so simple because the error terms can add up to a constant times $x$. At any rate, Estermann proved in 1931 that for any nonzero $k$ there are infinitely many $n$'s such that $n^2+k$ is square-free. I am sure the approach there can be easily modified to yield what you need. I found Estermann's paper in a more recent paper by Heath-Brown (arxiv.org/pdf/1010.6217.pdf). Estermann's paper is also online: digizeitschriften.de/dms/img/… – GH from MO Aug 5 '14 at 14:45

Here is a quick proof that the density in question exists and equals $$c:=\frac{2}{3}\prod_{p\geq 5}\left(1-\frac{2}{p^2}\right)\approx 0.553087\ .$$
Let $f(d)$ denote the number of solutions of the congruence $n(432n+1)\equiv 0\pmod{d}$. Note that, for $p$ prime, $f(p^2)=1$ when $p<5$ and $f(p^2)=2$ when $p\geq 5$. Let $P\geq 2$ be fixed. By a simple inclusion-exclusion sieve combined with the Chinese remainder theorem, we see that the number of $n\leq x$ such that $n(432n+1)$ is not divisible by the square of any prime $p\leq P$, equals $$x\prod_{p\leq P}\left(1-\frac{f(p^2)}{p^2}\right)+o_P(x)=x\prod_p\left(1-\frac{f(p^2)}{p^2}\right)+o_P(x)+O(x/P).$$ Observe that the infinite product on the right equals $c$. On the other hand, the number of $n\leq x$ such that $n(432n+1)$ is divisible by the square of some prime $p>P$, is at most $$\sum_{p>P}O(x/p^2)=O(x/P).$$ Altogether we see that the number of $n\leq x$ such that $n(432n+1)$ is square-free, equals $cx+o_P(x)+O(x/P)$. In particular, both the lower and the upper density of these numbers equals $c+O(1/P)$. These quantities are independent of $P$, hence upon letting $P\to\infty$, we see that they both equal $c$. So the density exists and also equals $c$.