The answer of following classical problem is surely known, but I can't find a reference



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> For which positive integer $n$ is the set $S\_n$ of primes of the form $x^2+n y^2$ ($x$, $y$ integers) determined by congruences?
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A set of prime $S$ is said *determined by congruences* if there is a positive integer $m$
and a set $A \subset (\mathbb{Z}/m\mathbb{Z})^\ast$ such that a prime $p$ not dividing $m$ is in $S$ if and only if $p$ modulo $m$ is in $A$. There is a natural place to look for this question: the book by Cox "primes number of the form $x^2+ny^2"$". Unfortunately I don't have it, my library doesn't have it, and I can't find it on the internet, except for some preview at Amazon and Google. From the table of content and the preview it seems that the book does not contain the answer to my question (otherwise I wouldn't ask) but it is still possible 
that the answer be hidden precisely in one of the sporadic pages that amazon doesn't 
want me to see. 


From that book one knows that a prime $p$ is in $S\_n$ if and only if it splits in the
ring class field $L\_n$ of the order $\mathbb{Z}[\sqrt{-n}]$ in the quadratic imaginary field $K\_n:=\mathbb{Z}[\sqrt{-n}]$. Therefore the question [becomes](https://mathoverflow.net/questions/11688/why-do-congruence-conditions-not-suffice-to-determine-which-primes-split-in-non-a): is $L\_n$ abelian over $\mathbb{Q}$? Now $H\_n:=Gal(L\_n/K\_n)$ is the ring class group of $\mathbb{Z}[\sqrt{-n}]$, hence abelian, and $Gal(L\_n/\mathbb{Q})$ is a semi-direct extension of $\mathbb{Z}/2\mathbb{Z}$ by $H\_n$, the action of the non-trivial element of $\mathbb{Z}/2\mathbb{Z}$ on $H\_n$ being $x \mapsto x^{-1}$. Hence, if I am not mistaken (am I?), the question is equivalent to



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> For which $n$ is the ring class group $H\_n$ killed by $2$?
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Thanks for any clue or reference. I am especially interested in the case $n=32$.



Primes of the form $x^2+ny^2$ and congruences.
You want the idoneal numbers,
<http://oeis.org/A000926> and <http://en.wikipedia.org/wiki/Idoneal_number>


See also pages 81-82 in Duncan A. Buell, *Binary Quadratic Forms*


Depending on what you mean by 32, the primes represented by $x^2 + 8 y^2$ are, in fact, given by congruences. However, half of those same primes are represented by $x^2 + 32 y^2,$ while the other half are represented by $4 x^2 + 4 x y + 9 y^2.$ The condition saying which are which is not simply congruences.


EDIT: it is possible 32 can be finished by biquadratic reciprocity, in which case it is in print somewhere. For instance, given a prime $p \equiv 1 \pmod 4,$ there is a representation $p = x^2 + 64 y^2$ in integers if and only if $2$ is a fourth power modulo $p.$ In comparison, we get $p = x^2 + 14 y^2$ for $p \neq 2,7 $ if and only if $ ( -14 | p ) = 1$ and $ (x^2 + 1)^2 \equiv 8 \pmod p$ has an integer solution (Cox page 115).


Either way, there is a monic irreducible polynomial $f\_{32}(z)$ of degree 4 (as $h(-128) = 4$) such that, if an odd prime $p$ does not divide the discriminant of $f\_{32}(z),$ then we can write $p = x^2 + 32 y^2$ if and only if $(-2 | p) = 1$ and $f\_{32}(z) \equiv 0 \pmod p$ has an integer solution. This is Cox, page 180, Theorem 9.2.


EDIT TOOO: What you want is Lemma 3.10 on page 333 of [LIU\_WILLIAMS](https://web.archive.org/web/20130620175150/http://zakuski.utsa.edu/%7Ejagy/Liu_Williams.pdf) Tamkang Journal of Mathematics, Volume 25, Number 4, Winter 1994.
I am going to need to check in various ways, but I already think that $p$ is represented by $x^2 + 32 y^2$ if and only if $p \equiv 1 \pmod 8$ and $(z^2 - 1)^2 + 1 \equiv 0 \pmod p$ has a solution with an integer $z.$ Checking... Yes, this is correct. Theorem 4.1 on the same page, Table right below it. There is a bit of work showing that one root and $p \equiv 1 \pmod 8$ actually shows four linear factors.


