Solvability of this General Pell's Equation

Question

If $p\equiv -1 \pmod{12}$ is a prime, does this general Pell's equation always have integer solutions for $x$ and $y$? $$x^2-3y^2=-p$$ I have tested that this is true for all values of $p$ up to $8\cdot 10^7$ but I failed to give a proof for it. It appears that there exists a solution with $x^2\le 2p$. It doesn't work for non-prime $p$, or simply $p \equiv -1 \pmod{3}$ or $p \equiv -1 \pmod{4}$. Thanks in advance.

Answer 1

Inspired by Dedekind's second proof for Fermat's sum of two squares theorem, we first note that $3$ is a quadratic residue modulo $p$ as by the law of quadratic reciprocity, $$\left(\frac{3}{p}\right)\left(\frac{p}{3}\right)={\left(-1\right)}^{\frac{p-1}{2}\frac{3-1}{2}}=-1.$$ Therefore, there exists an integer $m$ such that $m^2-3\equiv 0\pmod{p}$. Now consider the quadratic order $\mathbb{Z}[\sqrt{3}]$, an unique factorization domain. Note that $p$ is not a prime in $\mathbb{Z}[\sqrt{3}]$ since $p \mid m^2-3$ but $p \nmid m-\sqrt{3}$ and $p \nmid m-\sqrt{3}$. Hence, we can write $p=uv$ where $u=x_1+y_1\sqrt{3}\in \mathbb{Z}[\sqrt{3}]$ and $v=x_2+y_2\sqrt{3}\in \mathbb{Z}[\sqrt{3}]$ are not units (elements with norm 1). Note that $$p^2=N(p)=N(u)\cdot N(v)=(x^2_1-3y^2_1)(x^2_2-3y^2_2)$$ as the norm $N(\cdot)$ is a completely multiplicative function. Since neither of $u$ and $v$ is a unit, we have $N(u)=N(v)=-p$ (as $N(u)\equiv N(v)\equiv 0$ or $1 \pmod{3}$ but $p\equiv 2 \pmod{3}$ so $N(u),N(v)\neq p$) and thus $$-p=x^2_1-3y^2_1=x^2_2-3y^2_2$$ for some integers $x_1, x_2, y_1, y_2$. So the equation indeed has a solution for all $p\equiv 11 \pmod{12}$