Prove that the equation xy = 2006 (x + y) has integer solutions.

Let’s transform this equation:
xy = 2006 (x + y);
xy – 2006x – 2006y = 0;
Add 2006 ^ 2 to the left and right sides;
xy – 2006x – 2006y + 2006 ^ 2 = 2006 ^ 2;
Factor the left side to get the equation:
(y – 2006) (x – 2006) = 2006 ^ 2;
This equation has an integer solution, for example:
x = 4012; y = 4012 – is the solution to this equation.