N.T.TUAN wrote: For a given integer $d$, let us define $S = \{m^{2}+dn^{2}| m, n \in\mathbb{Z}\}$. Suppose that $p, q$ are two elements of $S$ , where $p$ is prime and $p | q$. Prove that $r = q/p$ also belongs to $S$ . Let $p=m_{0}^{2}+dn_{0}^{2}, k=\frac{m_{0}}{n_{0}}(mod \ p)=\sqrt{-d}(mod \ p).$ Let $q=pr=m^{2}+dn^{2}$. If $p|m, \ p|n$, then $r=(\frac{mm_{0}-nn_{0}d}{p})^{2}+d(\frac{nm_{0}+n_{0}m}{p})^{2}$. Else $m=\pm nk(mod \ p)$ and $r=(\frac{mm_{0}\pm nn_{0}d}{p})^{2}+d(\frac{nm_{0}\mp n_{0}m}{p})^{2}.$

following link is easy than what Rust posted http://www.mathlinks.ro/Forum/viewtopic.php?t=66248 but same . Sorry, because I don't remember it.