How to prove that there exist infinitely many primes of the form $3k-1, \ k\in \mathbb{N}$?
I know the proof that there exist infinitely many primes of the form $3k+1$, but I don't know how to do it in this case.
I tried to use:
Let $p_1,...,p_n$ be primes of the form $3k-1$. Then $N:=(3p_1\cdot \cdot \cdot p_n)-1$ is of the form $3k-1$ and $N$ must have at least one prime divisor of the form $3k-1$.
But $N$ is not divisible by any primes $p_1,...,p_n$.
So $N$ is a prime of the form $3k-1$ outside $p_1,...,p_n$.
This procedure can be continued indefinitely, so there exist infinitely many primes of the form $3k-1$.
Is this correct? Or is there something missing or wrong?
As 3 and 2 are coprime, by Dirichlet's theorem on arithmetic progressions there infinitely many prime numbers of the form 3*n+2, n>=0. By changing n by k-1, you obtain the desired result.