# Dirichlet's theorem on products of two primes in arithmetic progressions

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!

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.
• 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 $w_i:=v_i+u_ia+u_ibv \in \mathbb{N}$, $a+bw_i \in P^2$. – user237522 Oct 24 '18 at 10:57
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 $$k$$.