Example 1. If $m^{2}=m+1, n^{2}=n+1$, and $m \neq n$, then $m^{5}+n^{5}=$ $\qquad$ (Math Competition)
By analyzing, we compare $E_{1}$ with $x^{2}-x-1=0$ and find that $m, n$ are the two distinct real roots of the equation. By Vieta's formulas, we get
$$
\begin{aligned}
m+n= & 1, m n=-1 \\
m^{2}+n^{2} & =(m+n)^{2}-2 m n=3 \\
m^{3}+n^{3} & =(m+n)\left(m^{2}+n^{2}\right)-m n(m+n) \\
& =4 .
\end{aligned}
$$

By analogy with the above process, we can reduce the target expression to a lower-degree algebraic expression in terms of known values.
$$
\begin{array}{l} 
m^{4}+n^{4}=\left(m^{2}+n^{2}\right)^{2}-2 m^{2} n^{2} \\
=9-2 \times 1=7, \\
m^{5}+n^{5}=\left(m^{2}+n^{2}\right)\left(m^{3}+n^{3}\right)-m^{2} n^{2}(m+ \\
n)=11 .
\end{array}
$$