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Dirichlet's theorem on arithmetic progressions says that if $a$ and $b$ are coprime, then $\{a+bL\}_{L \in \mathbb{N}}$ contains infinitely many prime numbers.

I wonder if the following claim is true:

If $a$ and $b$ are coprime, then $\{a+bL\}_{L \in \mathbb{N}}$ contains infinitely many products of two prime numbers.

Remark: Please do not confuse my above claim with the following different claim: Denote the set of prime numbers by $P$. If $\gcd(a,b) \in P$, then $\{a+bL\}_{L \in \mathbb{N}}$ contains infinitely many products of two prime numbers. This claim is clearly true, since we can write $a=pa'$ and $b=pb'$, for some $p \in P$ and $\gcd(a',b')=1$. Then by Dirichlet's theorem, $\{a'+b'L\}_{L \in \mathbb{N}}$ contains infinitely many prime numbers, and then $\{a+bL\}=\{pa'+pb'L\}=\{p(a'+b'L)\}_{L \in \mathbb{N}}$ contains infinitely many products of two prime numbers.

Thank you very much!

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Sure. By Dirichlet, we can find infinitely many primes $p_i\equiv 1\pmod b$ and infinitely many primes $q_i\equiv a\pmod b$, in which case $p_i\times q_i$ works.

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  • $\begingroup$ I am adding the following explanation to your answer: $p_i=1+bu_i \in P$ and $q_i=a+bv_i \in P$, so $a+b(v_i+u_ia+u_ibv)=a+bv_i +bu_ia+bu_ibv_i=(1+bu_i)(a+bv_i)=p_iq_i \in P^2$. Therefore, for infinitely many $w_i:=v_i+u_ia+u_ibv \in \mathbb{N}$, $a+bw_i \in P^2$. $\endgroup$ – user237522 Oct 24 '18 at 10:57
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Take $p$ a prime factor of $a=pc$, and $L=pN$ for integers $N$, then there are infinitely many prime numbers of the form $c+bN$ and $p(c+bN)=pc+pbN=a+bL$ is a product of two primes.

If $a=\pm 1$ (so doesn't have a prime factor) take $(a+kb)+b(L-k)$ as an equivalent of the original form for some suitable $k$.

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  • $\begingroup$ Thank you very much! $\endgroup$ – user237522 Oct 24 '18 at 10:41

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