Problem 1826

The "only if" part is clear. We prove the "if" part. For $p=3$ one can take $(x, y)=(0,1)$. Assume $p \equiv 1$ $(\bmod 3)$. By quadratic reciprocity, $\left(\frac{-3}{p}\right)=\left(\frac{p}{3}\right)=1$. Thus $p$ splits in $\mathbb{Q}(\sqrt{-3})$. The ring of integers of $\mathbb{Q}(\sqrt{-3})$ is $\mathbb{Z}[\omega]$, where $\omega=\frac{-1+\sqrt{-3}}{2}$. Since $\mathbb{Z}[\omega]$ is a PID, there exists $\pi \in \mathbb{Z}[\omega]$ such that $N_{\mathbb{Q}(\sqrt{-3}) / \mathbb{Q}}(\pi)=p$. We claim that at least one of $\pi$, $\pi \omega$, and $\pi \omega^{2}$ belongs to $\mathbb{Z}[\sqrt{-3}]$ and thus is of the form $x+y \sqrt{-3}$ with $x, y \in \mathbb{Z}$. Taking norms, we then get $p=x^{2}+3 y^{2}$. To prove the claim, we may assume $\pi=\frac{a+b \sqrt{-3}}{2}$, where $a$ and $b$ are odd integers. Then either $4 \mid a-b$ (which is equivalent to $\pi \omega \in \mathbb{Z}[\sqrt{-3}]$ ) or $4 \mid a+b$ (which is equivalent to $\pi \omega^{2} \in \mathbb{Z}[\sqrt{-3}]$ ).