question about the representation of a prime $p$ by a quadratic form

Question

Do you know $2$ conditions for a prime $p$ to be represented by the form: $p=x^2+11y^2, \: \: x,y \in \mathbb{Z}$.

I know that one of the conditions is that $-11$ must be a quadratic residue modulo $p$.

Is it possible to express the second one in term of a monic polynomial $f_{11}(x)$ in $\mathbb{Z}[x]$ and divisible by $p$ ? Many thanks in advance.

Answer 1

$p=11$ in this represetaion is natural, as it is one of the ramifying primes in the extension $ \mathbb{Q} \subset \mathbb{Q}(\alpha)$, where $\alpha$ is a root of $x^3 − x^2 −x − 1$.

There is also a link to the Tribonacci numbers whereby for the n-th Tribonacci number, $T_n$ given by the recurrence $$T_{n+3} = T_{n+2}+T_{n+1}+T_n$$

with inital vlaues $T_0 = 0, T_1 = 1, T_2 =1$

then for primes $p$ not equial to $11$ or $19$, $T_{p-1}$ is divisible by $p$ if and only if $p=x^2+11y^2$ for $x, y \in \mathbb{Z}$

Answer 2

I found out that this is a very difficult problem that outreaches my mathematical abilities ! The quadratic discriminant of $x^2+11y^2$ is $D=-44$. The monic degree 3 polynomial I was searching for is $f_{11}(X)=X^3-2X^2+2X-2=\big(X-\mathfrak{f}(\tau_{q_0})\big)\big(X-\zeta^{-11}_{48}\mathfrak{f}_1(\tau_{q_1})\big)\big(X-\zeta^{11}_{48}\mathfrak{f}_2(\overline{\tau_{q_1})}\big)$

where $\mathfrak{f}, \mathfrak{f}_1, \mathfrak{f}_2$ are the Weber functions. I give very few explanations in case you want to form by yourself this polynomial.