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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?

Gerturter
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    That $N$ must have a least one prime divisor of the form $3k-1$ is true, but your proof should really include an explanation why it's true. – Barry Cipra Apr 26 '20 at 11:54

1 Answers1

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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.

Rémy Sigrist
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  • Why the down vote? – Johnduck Apr 26 '20 at 12:34
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    @johnduck I downvoted because it does not answer the OP's question (it is tagged "solution verification"), uses a very strong and non-trivial result from analytical number theory which is most likely not available here (it is tagged "elementary number theory") and does not use MathJax. – Marktmeister Apr 26 '20 at 14:43