4. Given that $m$ and $n$ are rational numbers, and the equation
$$
x^{2}+m x+n=0
$$

has a root $\sqrt{5}-2$. Then $m+n=$ $\qquad$ .
(2001, National Junior High School Mathematics Competition, Tianjin Preliminary Round)
$$
\begin{array}{l}
(\sqrt{5}-2)^{2}+m(\sqrt{5}-2)+n=0 \\
\Rightarrow(n-2 m+9)=(4-m) \sqrt{5} \\
\Rightarrow\left\{\begin{array} { l } 
{ n - 2 m + 9 = 0 , } \\
{ 4 - m = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
m=4, \\
n=-1
\end{array}\right.\right. \\
\Rightarrow m+n=3 \text {. } \\
\end{array}
$$