Infinite numbers of coprime pairs $k(6k-1)$.

Question

This question leads to the following consideration.

Are there infinite number of pairs of the form $\big(j(6j-1),\, k(6k−1)\big)$ where $j<k$ such that $j(6j-1)$ and $k(6k−1)$ are coprime? One sufficient condition is for both $k$ and $6k-1$ to be prime. This is different from Dirichlet's theorem on arithmetic progressions which does not specify $k$ has to be prime.

Answer 1

If $(a,b)=1$, then $(ab,c)=(a,c)(b,c)$. Since $(k,6k-1)=1$, it implies we want $$ 1=(k(6k-1),l(6l-1))=(k,l)(k,6l-1)(6k-1,l)(6k-1,6l-1), $$ hence $$ (k,l)=(k,6l-1)=(6k-1,l)=(6k-1,6l-1)=1. $$ There are infinitely many such $k,l$. For example put $l=k+1$ and the above conditions simplify to $5\nmid k$ and $7\nmid k+1$, so for example $k=35t+1,l=35t+2$ will do the trick.

Answer 2

Unfortunately, it's not yet been proven there are infinitely many $k$ and $6k - 1$ which are both prime (note this is a particular case of Dickson's conjecture). Instead, consider $k_i$ to be an infinite sequence of positive integers, starting with $k_1 = 3$, where for every pair of indices $i \neq j$, we have $\gcd(k_i(6k_i - 1), k_j(6k_j - 1)) = 1$. For $i \ge 1$, define

$$f(k_i) = k_i(6k_i - 1) \tag{1}\label{eq1A}$$

$$a_{i+1} = \prod_{j=1}^{i}f(k_j) \tag{2}\label{eq2A}$$

Also,

$$k_{i+1} = b_{i+1}a_{i+1} + 1 \tag{3}\label{eq3A}$$

where if $a_{i+1}$ is congruent to $1$, $3$ or $4$ mod $5$, then $b_{i+1} = 2$, else for $a_{i+1} \equiv 2 \pmod{5}$, then $b_{i+1} = 4$. This shows $k_{i+1}$ is coprime with $a_{i+1}$ so, from \eqref{eq2A}, also for all $f(k_j)$ for $1 \le j \le i$. In addition, we have

$$6k_{i+1} - 1 \equiv 6(b_{i+1}a_{i+1} + 1) - 1 \equiv 5 \pmod{a_{i+1}} \tag{4}\label{eq4A}$$

Note $5 \nmid f(k_1)$. For all $j \ge 1$, based on how $b_{j+1}$ is defined for \eqref{eq3A}, we have $\gcd(5,k_{j+1}) = 1$. Since by the choice of $b_{j+1}$ we also have $k_{j+1} \not\equiv 1 \pmod{5}$, then $6k_{j+1} - 1 \not\equiv 0 \pmod{5}$. Thus, from the definition in \eqref{eq2A} and using \eqref{eq1A}, we have $\gcd(a_{i+1}, 5) = 1$. Therefore, \eqref{eq4A} shows that $6k_{i+1} - 1$ is also coprime with $a_{i+1}$ and, thus, also for all $f(k_j)$ for $1 \le j \le i$.

We therefore have for all $i \ge 1$ that $f(k_{i+1})$ is coprime with all $f(k_{j})$ for $1 \le j \le i$.