On the equation \(x^ 2+Dxy+y^ 2=z^ k\) (Q1118636)

The author's result is the following. All integer solutions x,y,z of the equation \[ | x^ 2+Dxy+y^ 2| =z^ k \] where \(D=\pm 1,\pm 3,\pm 5,\pm 7,\pm 9\), \(k=1,2,3,...\), and \((x,y,z)=1\), \(3\nmid z\), are given by the formulae \[ | m^ 2+Dmn+n^ 2| =z,\quad \left( \begin{matrix} x\\ y\end{matrix} \begin{matrix} -y\\ x+Dy\end{matrix} \right)=\left( \begin{matrix} m\\ n\end{matrix} \begin{matrix} -n\\ m+Dn\end{matrix} \right)^ k\left( \begin{matrix} u\\ v\end{matrix} \begin{matrix} -v\\ u+Dv\end{matrix} \right), \] where m,n,u,v\(\in {\mathbb{Z}}\), \((m,n)=1\), \(| u^ 2+Duv+v^ 2| =1\). In the case \(D\neq \pm 7\), the assumption \(3\nmid z\) can be omitted.