Are $3$ and $19$ the only primes representable by the principal form with discriminant $-57$?

Question

I'm trying to see if there are other primes, but so far I only managed to get $3$ and $19$ by factoring $57$. How would I find other primes, if they do indeed exist?

Answer 1

As Will Jagy says, there is no form of discriminant $-57 = -3 \cdot 10$, so instead we will deal with discriminant $-4\cdot3\cdot19$. Moreover, it is hard to pitch this answer without knowing assumed knowledge, so I'll just write what I'd do. Let $K = \mathbb{Q}(\sqrt{-57})$.

We can compute the reducted BQFs of this discriminant as \begin{align*} &x^2 + 57y^2 & 3x^2 + 19y^2\\ &6x^2 + 6xy + 11y^2 & 2x^2 + 2xy + 29y^2 \end{align*} which all correspond to elements of order $2$ in the class group of $K$. Thus the Hilbert Class Field $L/K$ is an everywhere unramified abelian extension of $K$ with Galois group $C_2 \times C_2$. It is easy to check that $L = K(\sqrt{3}, \sqrt{19}) = \mathbb{Q}(i, \sqrt{3}, \sqrt{19})$.

A prime $p \neq 3, 19$ is represented by the principal form if and only if it splits in $L$. This is if and only if $-1$, $3$, and $19$ are squares modulo $p$.

The first is if and only if $p \equiv 1 \pmod 4$, so we assume that. For the second $$\left( \frac{q}{p} \right) = \left( \frac{p}{q} \right)$$ for any odd prime $q$ by our assumption on $p$ and quadratic reciprocity.

Thus $p \neq 3, 19$ is represented if and only if $p \equiv 1 \pmod 4$, $p \equiv 1 \pmod 3$ and $p \equiv 1, 4,5,6,7,9,11,16,17 \pmod{19}$.