, 2020

Solution:
We perform the linear transformation $(x, y) \rightarrow (x-y, x+y)$, which has the reverse transformation $(a, b) \rightarrow \left(\frac{a+b}{2}, \frac{b-a}{2}\right)$. Then the equivalent problem has a parabola has a vertical axis of symmetry, goes through $A=(0,200)$, a point $B=(u, v)$ in
$$S' = \{(x, y) \mid x+y>0, x>y, y<200, x, y \in \mathbb{Z}, \text{ and } x \equiv y \bmod 2\}$$
and a new vertex $W=(w, 0)$ on $y=0$ with $w$ even. Then $\left(1-\frac{u}{w}\right)^2=\frac{v}{200}$. The only way the RHS can be the square of a rational number is if $\frac{u}{w}=\frac{v'}{10}$ where $v=2\left(10-v'\right)^2$. Since $v$ is even, we can find conditions so that $u, w$ are both even:
$$\begin{aligned} v' \in \{1,3,7,9\} & \Longrightarrow (2 v')|u,\ 20|w \\ v' \in \{2,4,6,8\} & \Longrightarrow v'|u,\ 10|w \\ v'=5 & \Longrightarrow 2|u,\ 4|w \end{aligned}$$
It follows that any parabola that goes through $v' \in \{3,7,9\}$ has a point with $v'=1$, and any parabola that goes through $v' \in \{4,6,8\}$ has a point with $v'=2$. We then count the following parabolas:
- The number of parabolas going through $(2k, 162)$, where $k$ is a nonzero integer with $|2k|<162$.
- The number of parabolas going through $(2k, 128)$ not already counted, where $k$ is a nonzero integer with $|2k|<128$. (Note that this passes through $(k, 162)$.)
- The number of parabolas going through $(2k, 50)$ not already counted, where $k$ is a nonzero integer with $|2k|<50$. (Note that this passes through $\left(\frac{2k}{5}, 162\right)$, and any overlap must have been counted in the first case.)
The number of solutions is then
$$2\left(80+\frac{1}{2} \cdot 64+\frac{4}{5} \cdot 25\right)=264$$