Find all integer solutions to $x y=2003(x+y)$
We get

$$
\begin{gathered}
2003^{2}-2003 x-2003 y-x y=2003^{2} \\
(2003-x)(2003-y)=2003^{2}
\end{gathered}
$$

So $(2003-x)$ is a divisor of $2003^{2}$. As 2003 is a prime number, we conclude that $2003-x$ is $-2003^{2},-2003,-1,1,2003$, or $2003^{2}$, and $x$. Then for $x, y$ we obtain

$$
\begin{array}{ll}
x=2003^{2}+2003=4014012, & y=2004, \\
x=2003+2003=4006, & y=4006, \\
x=1+2003=2004, & y=4014012, \\
x=-1+2003=2002, & y=-401006, \\
x=-2003+2003=0, & y=0, \\
x=-2003^{2}+2002=-401006, & y=2002 .
\end{array}
$$