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 $$ \lim_{x\to \infty}\frac{\# \{n  \leq x\ : \ n \ \textrm{is squarefree integer, and } 432n+1 \ \textrm{is squarefree}\}}{x} = 1/2?$$ 
 A: 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 $c$.
For related comments see the introduction of this 1953 paper by Erdős.
