On the Integer Solutions of the Diophantine Equation \((x+y+z)^2=xyw\)

By using the theory of Pell equations, we prove that the Diophantine equation $(x+y+z)^2=xyw$ has infinitely many integer solutions. Moreover, we show that every positive integer solution of this Diophantine equation can generate infinitely many different positive integer solutions by the following transformation: \vspace{-2ex} \begin{figure}[H] \begin{center} \tikzstyle{format}=[rectangle,draw,thin,fill=white] \tikzstyle{test}=[diamond,aspect,draw,thin] \tikzstyle{point}=[coordinate,on grid,] \begin{tikzpicture} \node[] (T_1){$\left(x,y,z,w\right)$}; \node[point,right of=T_1,node distance=17mm](point1){}; \node[point,above of=point1,node distance=6mm] (point2){}; \node[point,below of=point1,node distance=6mm] (point3){}; \node[right of=point1,node distance=22mm] (T_2){$\left(x,wx-2x-2z-y,z,w\right),$}; \node[right of=point2,node distance=46mm] (T_3){$\left(x,y,kxyw-2x-2y-z,k^2xyw^2-2kw(x+y+z)+w\right),$}; \node[right of=point3,node distance=22mm] (T_4){$\left(wy-2y-2z-x,y,z,w\right),$}; \draw[-] (T_1.east)--(point1); \draw[-] (T_1.east)--(point2); \draw[-] (T_1.east)--(point3); \end{tikzpicture} \end{center} \end{figure} \vspace{-4ex} \noindent where $k>\frac{2}{x+y+z}$ and $k\in\mathbb{Q}^+$