3. Find a prime number $p$ greater than 2, such that the parabola
$$
y=\left(x-\frac{1}{p}\right)\left(x-\frac{p}{2}\right)
$$

has a point $\left(x_{0}, y_{0}\right)$ where $x_{0}$ is a positive integer and $y_{0}$ is the square of a prime number.
3. From the problem, we have
$$
\begin{array}{l}
y_{0}=\left(x_{0}-\frac{1}{p}\right)\left(x_{0}-\frac{p}{2}\right)=\frac{x_{0} p-1}{2} \cdot \frac{2 x_{0}-p}{p} . \\
\text { From } \frac{x_{0} p-1}{2}>\frac{2 x_{0}-p}{p}, \text { we know } \frac{2 x_{0}-p}{p}=1 .
\end{array}
$$

Therefore, $x_{0}=p, y_{0}=\frac{p^{2}-1}{2}$.
Given $y_{0}=q^{2}$ (where $q$ is a prime number), then
$$
\begin{array}{l}
p^{2}-1=2 q^{2} \Rightarrow(p+1)(p-1)=2 q^{2} \\
\Rightarrow 4|(p+1)(p-1), 4| 2 q^{2} \Rightarrow 2 \mid q .
\end{array}
$$

Thus, $q=2, p=3$.