N22 (22-3, Netherlands) Given integers $m, n$ satisfying $m, n \in\{1,2, \cdots, 1981\}$ and $\left(n^{2}-m n-m^{2}\right)^{2}=1$, find the maximum value of $m^{2}+n^{2}$.
If $m=n$, then $m=n=1$.
If $m \neq n$, from $n^{2}-n m-m^{2}= \pm 1$, we get $n>m$.
Since
$$
\begin{aligned}
\left(n^{2}-m n-m^{2}\right)^{2} & =\left((n-m)^{2}+m(n-m)-m^{2}\right)^{2} \\
& =\left(m^{2}-m(n-m)-(n-m)^{2}\right)^{2},
\end{aligned}
$$

thus, if $(m, n)$ satisfies the condition, then $(n-m, m)$ also satisfies the condition. Since $n>m$, we can start from $(n, m)$ and decrease to $(1,1)$. Conversely, starting from $(1,1)$, using $(n-m, m) \rightarrow(m, n)$, we can obtain all solutions satisfying $m, n \in\{1,2, \cdots, 1981\}$. That is,
$$
\begin{aligned}
& (1,1) \rightarrow(1,2) \rightarrow(2,3) \rightarrow(3,5) \rightarrow(5,8) \rightarrow(8,13) \rightarrow(13,21) \\
\rightarrow & (21,34) \rightarrow(34,55) \rightarrow(55,89) \rightarrow(89,144) \rightarrow(144,233) \\
\rightarrow & (233,377) \rightarrow(377,610) \rightarrow(610,987) \rightarrow(987,1597) .
\end{aligned}
$$

Therefore, the maximum value of $m^{2}+n^{2}$ is $987^{2}+1597^{2}=3524578$.