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Is there a nice elementary way to prove there is infinite prime numbers of form $5n+3$ (also for $5n+2$) with $n\in \mathbb{N}$?

I know how to do it for primes of form $pn+1$ for any prime $p\geq 3$ but not in this case.

I'm also familiar with the theorems of Schur and Murty.

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    $\begingroup$ This would be a good question for math.stackexchange. $\endgroup$ Commented Apr 29, 2020 at 7:55
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    $\begingroup$ The question mathoverflow.net/questions/16735/… is a more general version of this (and to my opinion suitable for this forum). It refers to kconrad.math.uconn.edu/blurbs/gradnumthy/dirichleteuclid.pdf for the results of Schur and Murty and more. The question mathoverflow.net/questions/15220/… is also relevant. $\endgroup$ Commented Apr 29, 2020 at 9:53
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    $\begingroup$ To add to the previous comment, here is link to the post on Mathematics: How to prove there is infinite prime numbers of form $5n+3$ without Dirichlet theorem? $\endgroup$ Commented Apr 29, 2020 at 12:04
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    $\begingroup$ In my opinion, this question would clearly be on-topic here, if it weren't for that Konstantinos's answer in Chris's link also answers this question. It is clearly not trivial with $5$ replaced with an arbitrary integer, evidenced by the fact that no such proof has appeared in the linked question (despite having 60 upvotes) and I personally consider the $m=5$ proof there also quite non-trivial. The question also demonstrates awareness of relevant literature, and I would say is of general interest to mathematicians (or at least to me)... but it is indeed arguably a duplicate. $\endgroup$
    – dhy
    Commented Apr 29, 2020 at 12:37
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    $\begingroup$ I wouldn't necessarily consider this a research-level question, but it's certainly not trivial. One can certainly adapt the proof of Dirichlet's Theorem to this special case (checking $L(\chi,s)\neq 0$ should be straightforward in this specific case), but this is probably not the kind of answer OP wants. $\endgroup$
    – Wojowu
    Commented Apr 29, 2020 at 12:40

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