Sets of Prime and Composite Numbers

Question

We know that all primes are of the form $ 6k ± 1 $ with the exception of 2 and 3.

We also know that not all numbers of the form $ 6k ± 1 $ are prime.

This leads to four distinct sets of pairs adjacent to a multiple of six:

1. Twin Primes, Example: $ 5, 7 $ (prime followed by a prime)
2. Twin Composites, Example: $ 119, 121 $ (composite followed by a composite)
3. Prime-Composite, Example: $ 23, 25 $ (prime followed by a composite)
4. Composite-Prime, Example: $ 35, 37 $ (composite followed by a prime)

The Twin Prime Conjecture states that there are infinitely many Twin Primes, but has yet to be proven.

Could it be proven that any of these four sets are infinite?

Answer 1

The composite-composite case is easy. By the Chinese remainder theorem there are infinitely many solutions of, for example, $$x\equiv0\pmod6\ ,\quad x\equiv1\pmod5\ ,\quad x\equiv-1\pmod7\ .$$ And for any such $x$ greater than $6$ we have $x-1,x+1$ are adjacent to a multiple of $6$, and $x-1$ is a multiple of $5$ and hence composite, and $x+1$ is a multiple of $7$ and hence composite.

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The composite-prime case follows from Dirichlet's theorem (which is not easy to prove). Consider the numbers $x=30k+6$. Then the numbers $x-1,x+1$ are adjacent to a multiple of $6$; and the numbers $x+1$ are prime infinitely often; and the numbers $x-1$ are always composite.

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Similarly, $x=30k-6$ covers the prime-composite case.

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And as you mentioned, the prime-prime case is still unsolved.

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Alternative proof for the composite case: consider the numbers $$x=6\times119\times121k+120\ .$$ Then $x-1$ is always a multiple of $119$, and $x+1$ is always a multiple of $121$.

Or to keep the numbers a bit smaller, $x=6\times7\times11k+120$.

Answer 2

It is very simple to construct an infinite sequence for the Prime-Composite case:

$$23+60n,25+60n$$

$25+60n$ will generate an infinite amount of composite numbers, all of which are divisible by $5$ (in fact, it will generate only composite numbers).

$23+60n$ will generate an infinite amount of prime numbers, since $23$ and $60$ are coprime integers (according to Dirichlet's theorem on arithmetic progressions).

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It is very simple to construct an infinite sequence for the Composite-Prime case:

$$35+60n,37+60n$$

$35+60n$ will generate an infinite amount of composite numbers, all of which are divisible by $5$ (in fact, it will generate only composite numbers).

$37+60n$ will generate an infinite amount of prime numbers, since $37$ and $60$ are coprime integers (according to Dirichlet's theorem on arithmetic progressions).

Answer 3

Case $2$ is answered so I will address the others. For case $3$, by Dirichlet's theorem there are an infinite number of primes of the form $10 \cdot n + 3$, and $10 \cdot n + 5$ is always divisible by $5$. There is a similar example for case $4$.