I'm trying to figure out a way to prove this:
Given arithmetic progression $ax + b$ where $a$, $b$ coprime and $ax + b$ is prime infinitely often, it is the case at least once that $ax + b - 2$ is semiprime when $ax + b$ is prime.
I'm looking for a way either to prove it, or whether an existing theorem is already known from which this would follow.
Stated another way, given arithmetic progression $ax + b$ where $a, b$ coprime and $ax + b$ is prime infinitely often, it is at least once the case that $ax + b$ produces a $cd$ + 2 prime, where c, d are prime. The list of $cd$ + 2 primes is A063638 on OEIS.
What we know: A Chen prime is a prime $p$ such that $p + 2$ is either prime or semiprime. Chen proved in 1966 there are infinitely many of these. There are also infinitely many primes $p$ such that $p - 2$ is specifically semiprime (not just prime-or-semiprime). Friedlander and Iwaniec proved this in their 2010 book.
Note that I'm not asking whether $ax + b - 2$ is semiprime infinitely often. I just need it to happen once, that $ax + b$ is prime while $ax + b - 2$ is semiprime.
A more specific version of the problem is this:
Given arithmetic progression $P(P + 2)x + (P + 4)$ where $P$, $P + 2$ are twin primes, it is the case at least once that $P(P + 2)x + (P + 2)$, i.e., $(Px + 1)(P + 2)$ is semiprime when $P(P + 2)x + (P + 4)$ is prime.
The deeper problem I'm trying to show is that it is the case at least once that $Px + 1$ is prime when $P(P + 2)x + (P + 4)$ [or, said another way, $(Px + 1)(P + 2) + 2$] is also prime. I'm trying to use Chen's theorem as a strategy to show this--- in other words, to show that $(Px + 1)(P + 2)$ must be semiprime at least once (and so, ergo, $Px + 1$ must be prime at least once) when $P(P + 2)x + (P + 4)$ is prime. All my efforts to prove one just leads to a tautology where I must assume the other first.
Any thoughts, suggestions or help would be appreciated.
2 and 105 are coprime, so there are infinitely many primes of the form $105x+2$, but there are no semiprimes of the form $105x+2-2$ since $105=3\cdot5\cdot7.$
For your more specific problem, given some prime $P$ with $P+2$ also prime, you're looking for some $x$ with $Px+1$ and $P(P+2)x+(P+4)$ both prime. These are both linear, so this is a special case of Dickson's conjecture. I don't believe any special cases are known.