3. If $m^{2}=m+1, n^{2}=n+1$, and $m \neq n$, then $m^{5}+n^{5}=$ $\qquad$
(Fourth Jiangsu Province Junior High School Mathematics Competition)
Given that $m$ and $n$ are the roots of the equation $x^{2}=x+1$, i.e., $x^{2}-x-1=0$.
Let $a_{k}=m^{k}+n^{k}$. Then
$$
a_{k}=a_{k-1}+a_{k-2}(k \geqslant 3) \text {. }
$$

By $a_{1}=m+n=1$,
$$
a_{2}=(m+n)^{2}-2 m n=3 \text {, }
$$

we know $a_{3}=a_{2}+a_{1}=4, a_{4}=a_{3}+a_{2}=7$.
Therefore, $m^{5}+n^{5}=a_{5}=a_{4}+a_{3}=11$.