12.37 In the Cartesian coordinate system $x-o-y$, we color the points with natural number abscissas and perfect square ordinates (squares of natural numbers) as red points. Determine all the "red points" through which the graph of the function $y=(x-36)(x-144)-1991$ passes.
(China Beijing High School Grade 1 Mathematics Competition, 1991)
【Sol】Let $y=m^{2}, m \in \mathbb{N}$, then
$$
\begin{array}{c}
(x-36)(x-144)-1991=m^{2}, \\
x^{2}-180 x+3193=m^{2}, \\
(x-90)^{2}-4907=m^{2} .
\end{array}
$$

Let $x-90=k$, then

Therefore,
$$
\begin{array}{c}
k^{2}-m^{2}=4907=7 \cdot 701, \\
(k+m)(k-m)=7 \cdot 701 .
\end{array}
$$

Since 7 and 701 are both prime numbers, and
$$
k-m<k+m \text {, }
$$

we can only get the following four systems of equations.
$\mathrm{I} \cdot\left\{\begin{array}{l}k-m=1, \\ k+m=4907 .\end{array}\right.$
II . $\left\{\begin{array}{l}k-m=7, \\ k+m=701 .\end{array}\right.$
III. $\left\{\begin{array}{l}k-m=-4907, \\ k+m=-1 .\end{array}\right.$
$\mathbb{V} \cdot\left\{\begin{array}{l}k-m=-701, \\ k+m=-7 .\end{array}\right.$

Solving the systems of equations III and IV yields $x$ as a negative integer, which does not meet the problem's requirements.
Solving the systems of equations $\mathrm{I}$, II, we get
$$
\left\{\begin{array}{l}
k=2454, \\
m=2453 .
\end{array} \text { or } \left\{\begin{array}{l}
k=354, \\
m=347 .
\end{array}\right.\right.
$$

Thus, $x=2544$ or $x=444$, and the corresponding $y=6017209$, or $y=120409$. Therefore, the graph of $y=(x-36)(x-144)-1991$ only passes through the points $(2544, 6017209)$, $(444, 120409)$.