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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 $k\in P$?

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    $\begingroup$ 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 $a,d$. $\endgroup$ – Wojowu Mar 6 at 14:31
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    $\begingroup$ See Dickson's conjecture. $\endgroup$ – Robert Israel Mar 6 at 15:20
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    $\begingroup$ 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.) $\endgroup$ – GH from MO Mar 6 at 16:08

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