28th Hellenic Mathematical Olympiad

After manipulation equation becomes $$\begin{array}{rl}{x}^2{y}^2(x-y{)}^2-6^2& =0,x,y\in \mathbb{Z}\\ & \iff [xy(x-y)-6][xy(x-y)+6]=0\\ & \iff xy(x-y)=6,x,y\in \mathbb{Z}\text{ or }xy(x-y)=-6,x,y\in \mathbb{Z}\\ & \iff xy(x-y)=6,x,y\in \mathbb{Z}\text{ (1) or }xy(y-x)=6,x,y\in \mathbb{Z}\text{ (2)}\end{array}$$ --- From (1) and (2) it is clear that, if $(x_0,y_0)$ is a solution of (1), then the pair $(y_0,x_0)$ is a solution of (2) and vice versa. Thus is enough to solve equation (1). Since $x,y\in \mathbb{Z}$, equation (1) is equivalent to: $$\begin{array}{l}\{xy=6,x-y=1\}({\Sigma }_1)\text{ or }\{xy=-6,x-y=-1\}({\Sigma }_2)\\ \text{or }\{xy=3,x-y=2\}({\Sigma }_3)\text{ or }\{xy=-3,x-y=-2\}({\Sigma }_4)\\ \text{or }\{xy=1,x-y=6\}({\Sigma }_5)\text{ or }\{xy=-1,x-y=-6\}({\Sigma }_6)\\ \text{or }\{xy=2,x-y=3\}({\Sigma }_7)\text{ or }\{xy=-2,x-y=-3\}({\Sigma }_8).\end{array}$$ From the eight systems only $({\Sigma }_1),({\Sigma }_3),({\Sigma }_8)$ have integer solutions: $$\begin{array}{rl}(x,y)& =(3,2),(x,y)=(-2,-3),(x,y)=(3,1),\\ (x,y)& =(-1,-3),(x,y)=(-2,1),(x,y)=(-1,2).\end{array}$$ According to the previous discussion equation (2) has also the solutions: $$\begin{array}{rl}(x,y)& =(2,3),(x,y)=(-3,-2),(x,y)=(1,3),\\ (x,y)& =(-3,-1),(x,y)=(1,-2),(x,y)=(2,-1).\end{array}$$