Determine the maximum value of $m^{2}+n^{2}$, where $m, n \in\{1,2, \cdots, 1981\}$, and $\left(n^{2}-\right.$ $\left.m n-m^{2}\right)^{2}=1$.

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If $m=n$, from $\left(n^{2}-m n-m^{2}\right)^{2}=1$ we know $m=n=1$. If $m \neq n$, then from $n^{2}-$ $m n-m^{2}= \pm 1$ we get $n>m$. Let $n=m+u_{k}$, then $\left[\left(m+u_{k}\right)^{2}-m\left(m+u_{k}\right)-m^{2}\right]^{2}=1$, simplifying to $\left(m^{2}-u_{k} m-u_{k}^{2}\right)^{2}=1$. Thus, $m>u_{k}$. Let $m=u_{k}+u_{k-1}$, similarly we have $\left(u_{k}^{2}-u_{k-1} u_{k}-u_{k-1}^{2}\right)^{2}=1$.

If $u_{k}+u_{k-1}$, the above steps can continue, until $\left(u_{k-l-1}^{2}-u_{k-l-1} \cdot u_{k-l}-u_{k-l}^{2}\right)^{2}=$ 1, and $u_{k-l}=u_{k-l-1}=1$. This results in the sequence: $n, m, u_{k}, u_{k-1}, \cdots, u_{k-l}, u_{k-l-1}$. Any three consecutive terms in this sequence satisfy $u_{i}=u_{i-1}+u_{i-2}$, which is exactly a Fibonacci-like sequence. The Fibonacci numbers in $\{1,2, \cdots, 1981\}$ are: $1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597$. It is clear that when $m=987, n=1597$, $m^{2}+n^{2}-3524578$ is the maximum value that satisfies the condition.