A strengthening of Dirichlet prime number theorem

Dirichlet Theorem on arithmetic progression states that the sequence $$\{a+kd\}_{k=1}^{\infty}$$ contains infinitely many primes when $$(a,d)=1$$. In other words if we let $$A=\{a+kd\}_{k=1}^{\infty}$$ and $$P$$ is the set of primes then $$|A \cap P|=\infty$$. My question is about a subset of this sequence. Let $$B=\{a+kd \in A \ | \ k\in P,\ a+d\equiv 1 \!\!\! \mod 2 \}$$. Is it still true that $$|B \cap P|=\infty$$? If not, what sort of condition / refinement would assure that such a subset has infinite cardinality without compromising the condition $$k\in P$$?

• The answer is no, as for $a,d$ odd, $a+kd$ is even for all $2\neq k\in P$. Excluding this case, this is an open problem for all $a,d$. – Wojowu Mar 6 at 14:31
• – Robert Israel Mar 6 at 15:20
• A more appropriate title would be: a strengthening of Dirichlet's prime number theorem. (Special case means a weaker statement. A special case of a theorem is still a theorem, not an unsolved problem.) – GH from MO Mar 6 at 16:08