29. How may pairs of integers $(x, y)$ satisfy the equation
$$
\sqrt{x}+\sqrt{y}=\sqrt{200600} ?
$$
29. Ans: 11 .
Since $x=200600+y-20 \sqrt{2006 y}$ is an integer, $y=2006 u^{2}$ for some positive integer $u$. Similarly $x=2006 v^{2}$ for some positive integer $v$. Thus $u+v=10$. There are 11 pairs in total, namely $(0,10),(1,9), \ldots,(10,0)$.