Example 4. If $m^{2}=m+1, n^{2}=n+1$, then what is the value of $m^{5}+n^{5}$?
(1989, Jiangsu Province High School Mathematics Competition)
Let $S_{k}=m^{k}+n^{k}$, construct the recurrence relation for $S_{k}$.
$$
\begin{array}{l}
\because m^{2}=m+1, n^{2}=n+1, \\
\therefore m^{k}=m^{k-1}+m^{k-2}, n^{k}=n^{k-1}+n^{k-2} .
\end{array}
$$

Adding the two equations gives $S_{k}=S_{k-1}+S_{k-2}$.
$$
\begin{aligned}
\therefore S_{5} & =S_{4}+S_{3}=\left(S_{3}+S_{2}\right)+\left(S_{2}+S_{1}\right) \\
& =\cdots=3 S_{2}+2 S_{1} .
\end{aligned}
$$

From the given conditions, $S_{2}=m+n-2$, and $m, n$ are the roots of the quadratic equation $x^{2}-x-1=0$, so $m+n=1$, i.e., $S_{1}=1, S_{2}=1+2=3$.
Thus, $S_{5}=3 \times 3+2 \times 1=11$.
Note: Here, we first construct the recurrence relation for $S_{k}=m^{2}+n^{k}$, then express $S_{5}$ in terms of $S_{1}$ and $S_{2}$, and solve the problem. This is a general method for solving problems of the form $a^{n}+b^{n}$.