30 Find the value of the smallest positive integer $m$ such that the equation
$$
x^{2}+2(m+5) x+(100 m+9)=0
$$
has only integer solutions.
30 Answer: (90)
$x^{2}+2(m+5) x+(100 m+9)=0 \Rightarrow x=-(m+5) \pm \sqrt{(m-45)^{2}-2009}$. This yields integer solutions if and only if $(m-45)^{2}-2009$ is a perfect square, say $n^{2}$.
Hence $(m-45)^{2}-n^{2}=2009=7^{2} \times 41 \Rightarrow$
$$
\begin{array}{c}
|m-45|+n=2009 \text { and }|m-45|-n=1 \text {, or } \\
|m-45|+n=287 \text { and }|m-45|-n=7 \text {, or } \\
|m-45|+n=49 \text { and }|m-45|-n=41 .
\end{array}
$$

Solving, $n=45 \pm 1005,45 \pm 147,45 \pm 45 \Rightarrow$ The smallest positive $n$ is 90 .