Let $\{u_n\}_{n\geq 0}$ be an arithmetic sequence defined by $u_n=an+b \quad (a,b\in\mathbb{N}^*)$.
a) Let $a=6$. Give 2 values of $b\leq 6$ s.t for each such value of $b$, $\{u_n\}$ has infinite primes.
b) Let $a=10$. Prove that there are infinite values of $b$ s.t for each such value of $b$, $\{u_n\}$ has infinite primes.
I have done part a) so far (indeed there are infinite primes of the form $6k+1, 6k+5$).
I'm stuck at part b). I don't want to use Dirichlet's theorem for this problem. Could anyone have any idea to tackle this problem without mentioning Dirichlet's theorem?
If there is a single value of $b$ then the family of $10k + b$ also satisfies, so one only need to prove that there is one such b.
Let $S=\{0,1,2,3,..,9\}$. Let $N_b=\{n\in \mathbb{N}\mid n=10k+b,b\in S,k\in \mathbb{N}\}$ be the set of natural number in the sequence $10k+b$, $P$ is the set of primes, $P_b = P \cup N_b$ is the set of primes in sequence $10k+b$.
Short proof: Assume that $\left| P_b\right|$ is finite for each $b$, then we have $$\sum{\left| P_b\right|} \geq \left| \bigcup{P_b}\right| = \left| \bigcup{P\cap N_b}\right| = \left| P \cap \mathbb{N}\right|=|P|$$
But this means there are finitely many primes, which is wrong, so there is at least one $b$ such that the set $P_b$ is infinite $\square$
Explanation: since every prime must belong to one and only one of the sets $P_b$, the sum of their sizes must equal to the number of primes. Therefore at least one of them must be infinite, hence it completes the proof