2. In the Cartesian coordinate system, points of the form $\left(m, n^{2}\right)$ are painted red (where $m, n$ are integers), referred to as red points, and their surrounding points are not colored. Then, the parabola $y=x^{2}-$ $196 x+9612$ has $\qquad$ red points.
2. 2 .

Let $\left(m, n^{2}\right)$ be on the parabola $y=x^{2}-196 x+9612$, then $n^{2}=m^{2}-196 m+9612$.
Completing the square and factoring, we get
$$
\begin{array}{l}
(n+m-98)(n-m+98) \\
=8=2 \times 4=(-2) \times(-4) .
\end{array}
$$
$\because m, n$ are integers,
$\therefore n+m-98$ and $n-m+98$ are of the same parity.
When $n+m-98$ takes $-4,-2,2,4$ in turn, $n-m+98$ takes $-2,-4,4,2$ in turn. Therefore, $\left(m, n^{2}\right)=(97,9)$ or $(99,9)$.