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Is there an integer $n\ge1$ such that every prime $p\equiv1\pmod{9}$ is representable in the form $x^2+ny^2$?

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    $\begingroup$ Set $n$ equal to $p-1$ and set $x$ and $y$ equal to $1$. Is that what you are asking? $\endgroup$
    – Jason Starr
    Commented Dec 11, 2017 at 12:04
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    $\begingroup$ I imagine, Jason, that OP doesn't want $n$ to depend on $p$. Perhaps OP could edit the question, to clarify this point? $\endgroup$
    – Gerry Myerson
    Commented Dec 11, 2017 at 12:06
  • $\begingroup$ @JasonStarr, Thanks for your comment, I am trying to find a representation of $p$ in the form $x^2+ny^2$ with $(x,y)=1$. $\endgroup$
    – square-free
    Commented Dec 11, 2017 at 12:06
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    $\begingroup$ Have you read David Cox's book *Primes of the form $X^2+nY^2$"? $\endgroup$
    – Joe Silverman
    Commented Dec 11, 2017 at 12:17
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    $\begingroup$ Is this a homework problem or take-home exam question? What is the motivation for asking about it here? There is such an $n$ that works, and this is a standard exercise in the initial parts of algebraic number theory. As a hint, there is an $n$ which even works for all $p\equiv 1 \bmod 3$. $\endgroup$
    – nfdc23
    Commented Dec 11, 2017 at 13:02

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Yes, $n=3$ works, and no other integer works. The "no other" part follows easily by inspecting $p=19$ and $p=37$. The "works" part was conjectured by Fermat in 1654 and proved by Euler in 1772. In modern terminology, the proof is quite simple, and it goes as follows (I give an outline):

A prime $p\nmid 2n$ is represented by a primitive form of discriminant $-4n$ if and only if $(-n/p)=1$ (see Corollary 2.6 in Cox: Primes of the form $x^2+ny^2$). In particular, if $x^2+ny^2$ is the only primitive form of discriminant $-4n$ up to equivalence, then a prime $p\nmid 2n$ is represented by it if and only if $(-n/p)=1$. This is certainly the case for $n=3$, and then also $(-n/p)=(p/3)$ by quadratic reciprocity. So we get that a prime $p>3$ is represented by $x^2+3y^2$ if and only if $p\equiv 1\pmod{3}$.

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