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The question is

Are there infinitely many integers $n$ such that $x^2 + ny^2$ represents infinitely many primes?

It is known that (Introductory Algebraic Number Theory by S.Alaca and K.Williams)

If $\mathbb{Z}[\sqrt d]$ is a principal ideal domain for some squarefree integer $d$ and $q$ be an odd prime such that $$(d\mid q) =1 $$ Then there exist integers $u$ and $v$ such that $$q = \mid u^2 - dv^2 \mid $$

Here $(.|.)$ is Legendre's symbol.

I think there is a conjecture that there are infinitely many squarefree positive integers $d$ such that $\mathbb{Z}[\sqrt d]$ is a PID. If this conjecture is true then there will be infinitely many integers $n$ such that $x^2 + ny^2$ represents infinitely many primes using the above theorem.

Is there a simple criterion other than the above conjecture which I am not seeing? It is sufficient even if you provide the references.

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  • $\begingroup$ How did you get from $|u^2-nv^2|$ to $x^2+ny^2$? Won't you want to work with $\Bbb Z[i\sqrt n]$ for that? $\endgroup$ Oct 27, 2021 at 12:04
  • $\begingroup$ I understand that there is confusion in the notation, I am actually considering all integers $n$ in the question $\endgroup$ Oct 27, 2021 at 12:11
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    $\begingroup$ Let $n\equiv 1\pmod 4$ be a square-free integer. There exist infinitely many primes $p$ with $p=x^2+ny^2$ if and only if there exist infinitely many primes $p$ which split as a product of principal ideals in $\mathbb{Z}[\sqrt{-n}]$. By Kummer-Dedekind, a prime $p\neq 2$ splits iff $-n$ is a non-zero quadratic residue mod $p$. With quadratic reciprocity, it's easy to show there exist infinitely many such primes. The problem is that the primes above $(p)$ must be principal too. PIDs are a special case because they have class number $1$. $\endgroup$
    – Mastrem
    Oct 27, 2021 at 12:27
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    $\begingroup$ Probably an answer in math.toronto.edu/~ila/Cox-Primes_of_the_form_x2+ny2.pdf $\endgroup$ Oct 27, 2021 at 12:39
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    $\begingroup$ Every primitive binary quadratic form represents infinitely many primes. This is due to Dirichlet. $\endgroup$
    – user23365
    Oct 27, 2021 at 13:02

1 Answer 1

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A theorem due to Dirichlet (completely proved only later, for example by Weber here) states that every primitive binary quadratic form represents infinitely many primes. The proof is much simpler than the theorem on primes in arithmetic progressions and uses the fact that $(s-1) \zeta_K(s)$ has a nonzero limit as $s \to 1$.

The result can also be deduced from the known fact that each ideal class contains infinitely many prime ideals.

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