Using elementary methods to prove infinitely many primes mod n

Question

I was reading an elementary number theory text looking to enhance my knowledge and I came across the relatively simple task of proving there existed infinitely many primes of the form $4k-1$ (of course, without Dirichlet). My very elementary proof is as follows:

Assume there exist only $n$ finitely many such primes: then let $m=4(p_1p_2\cdots p_n)-1$. This is a (odd) number of the form $4k-1$ and thus must have factors of form $4k-1$, for otherwise the number would be of the form $4k+1$.

Is there such a simple generalization of this proof? I can see that this proof does not work for some, such as the $4k+1$ case found here. For instance, please provide a similar proof that there exists infinitely many primes of the form $15k+4$ (randomly chosen numbers). Thanks.

Answer 1

This question has been asked many times. These are called Euclidean proofs of special cases of Dirichlet's theorem on primes in arithmetic progressions. Keith Conrad has a nice article on it, which includes a complete characterization of when such a proof exists. Thanks to the characterization, since $$4^2\equiv 1\pmod{15},$$ there exists such a proof in the case that you requested, though I don't know what explicit polynomial would be used.

I wrote about this general problem in this post as well.

By the way, a Euclidean proof does exist in the $1\pmod 4$ case. You can use the polynomial $n^2+1,$ but you need to include $2$ in the product of the presumed finite list of prime to get the contradiction.

EDIT: I found a polynomial for $15k+4$ with proof in pages 92-94 of Problems in Algebraic Number Theory. It is $$n^4-n^3+2n^2+n+1.$$ The brother of one of the authors recommended the book to me years ago, but it never suited me... finally found use for it.