Example 14. If $m^{2}=m+1, n^{2}=n+1$, and $m \neq$ $n$, then $m^{5}+n^{5}=$ $\qquad$
$$
\begin{array}{l}
\text { Sol } \because(a-b)\left(a^{n}+b^{n-1}\right) \\
=a^{n}+b^{n}+a b\left(a^{n-2}+b^{n-2}\right), \\
\therefore a^{n}+b^{n}=(a+b)\left(a^{n-1}+b^{n-1}\right) \\
-a b\left(a^{n-2}+b^{n-2}\right) \text {. } \\
\end{array}
$$

Let $S_{n}=a^{n}-b^{n}$, we get the recursive formula
$$
S_{n}=(a+b) S_{n-1}-a b S_{n-2} .(n=2,3, \cdots)
$$

From the given, $m, n$ are the unequal real roots of $x^{2}-x-1=0$, hence
$$
\begin{array}{l}
m+n=1, m n=-1, m^{2}+n^{2}=3 . \\
\text { Therefore, } S_{n}=S_{n-1}+S_{n-2} . \\
\text { Thus, } S_{3}=S_{2}+S_{1}=3+1=4, \\
S_{4}=S_{3}+S_{2}=4+3=7, \\
S_{5}=S_{4}+S_{3}=7+4=11 .
\end{array}
$$

Therefore, $m^{5}+n^{5}=11$.
$$