12. A. In the Cartesian coordinate system $x O y$, points with integer abscissas and ordinate values that are perfect squares are called "good points". Find the coordinates of all good points on the graph of the quadratic function $y=(x-90)^{2}-4907$.
12. A. Let $y=m^{2},(x-90)^{2}=k^{2}(m, k$ are non-negative integers). Then,
$$
k^{2}-m^{2}=7 \times 701=1 \times 4907,
$$

i.e., $(k-m)(k+m)=7 \times 701=1 \times 4907$.
$$
\text { Then }\left\{\begin{array} { l } 
{ k + m = 7 0 1 , } \\
{ k - m = 7 ; }
\end{array} \left\{\begin{array}{l}
k+m=4907, \\
k-m=1 .
\end{array}\right.\right.
$$

Solving, we get $(k, m)=(354,347),(2454,2453)$.
Therefore, there are 4 good points, with coordinates
$$
\begin{aligned}
(x, y)= & (444,120409),(-264,120409), \\
& (2544,6017209),(-2364,6017209) .
\end{aligned}
$$