Twin square-free numbers of the form $6k-1,6k+1$?

Question

Is it easy to show (or even known) that there are infinitely many square-free pairs $6k-1,6k+1$?

(Presumably, not disproven yet, since a lot of people would be wasting their time on the twin prime conjecture if it was.)

Answer 1

Given a prime $p > 3$, the setf $A_p = \{k \in \mathbb{N}\setminus \{0\} : 6k \equiv \pm 1 \pmod{p^2}\}$ contains exactly two elements in each interval of length $p^2$.

Also, if $n\geq 7$ then $A_p\cap \{1,2,\dots,n\}=\emptyset$ when $p\geq n$, since $p^2\mid 6k\pm 1$ means $6k+1\geq p^2$ which in turn means $k\geq\frac{p^2-1}{6} \geq p+1>n$.

Thus

$$\biggl\lvert\{ 1,\dotsc,n\} \cap \bigcup_{p > 3} A_p\biggr\rvert \leqslant \sum_{3 < p \leqslant n} 2\biggl\lceil \frac{n}{p^2}\biggr\rceil \leqslant 2\pi(n) + 2n\sum_{3 < p\leqslant n} \frac{1}{p^2},$$

and since $\frac{\pi(n)}{n}\to 0$, the natural density of

$$A = \bigcup_{p > 3} A_p$$

is bounded above by

$$\sum_{ p > 3} \frac{2}{p^2} < 2\sum_{k = 2}^\infty \frac{1}{(2k+1)^2} < 1.$$

Hence the natural density of $B = \mathbb{N}\setminus (A\cup \{0\})$ is positive, in particular $B$ is infinite.

(Corrected argument per Thomas Andrews' comment.)