12. (NET) $)^{1 \mathrm{moa}}$ Determine the maximum value of $m^{2}+n^{2}$ where $m$ and $n$ are integers satisfying
$$
m, n \in\{1,2, \ldots, 100\} \quad \text { and } \quad\left(n^{2}-m n-m^{2}\right)^{2}=1
$$
12. We will solve the contest problem (in which $m, n \in\{1,2, \ldots, 1981\}$ ). For $m=1, n$ can be either 1 or 2 . If $m>1$, then $n(n-m)=m^{2} \pm 1>0$; hence $n-m>0$. Set $p=n-m$. Since $m^{2}-m p-p^{2}=m^{2}-p(m+p)=$ $-\left(n^{2}-n m-m^{2}\right)$, we see that $(m, n)$ is a solution of the equation if and only if $(p, m)$ is a solution too. Therefore, all the solutions of the equation are given as two consecutive members of the Fibonacci sequence
$$
1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584, \ldots .
$$

So the required maximum is $987^{2}+1597^{2}$.