Example 5 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. [5]
(2010, I Love Mathematics Junior High School Summer Camp Mathematics Competition)
Let $y_{0}=t^{2}$ (where $t$ is a prime number). Then
$$
\left(x_{0}-\frac{1}{p}\right)\left(x-\frac{p}{2}\right)=t^{2} \text {, }
$$

which implies $\left(p x_{0}-1\right)\left(2 x_{0}-p\right)=2 p t^{2}$.
Since $p$ is a prime number and $p \times\left(p x_{0}-1\right)$, we have
$$
p\left|\left(2 x_{0}-p\right) \Rightarrow p\right| 2 x_{0} \text {. }
$$

Since $p>2$, $p$ and 2 are coprime. Therefore, $p \mid x_{0}$.
Let $x_{0}=k p\left(k \in \mathbf{N}_{+}\right)$.
Substituting into equation (1) gives
$$
\begin{array}{l}
\left(k p^{2}-1\right)(2 k p-p)=2 p t^{2} \\
\Rightarrow\left(k p^{2}-1\right)(2 k-1)=2 t^{2} \\
\Rightarrow(2 k-1) \mid 2 t^{2} .
\end{array}
$$

Since $2 k-1$ and 2 are coprime, we have $(2 k-1) \mid t^{2}$.
Since $t$ is a prime number, $2 k-1$ can only be $1, t, t^{2}$.
If $2 k-1 \geqslant t$, then
$$
k p^{2}-1 \geqslant 3^{2} k-1=9 k-1>4 k-2=2 t \text {. }
$$

Thus, $\left(k p^{2}-1\right)(2 k-1)>2 t^{2}$, which contradicts equation (2).
Therefore, $2 k-1<t \Rightarrow 2 k-1=1 \Rightarrow k=1$.
Substituting $k=1$ into equation (2) gives
$$
p^{2}-1=2 t^{2} \Rightarrow p^{2}=2 t^{2}+1 \text {. }
$$

If $t=3$, then $p^{2}=2 \times 3^{2}+1=19$, and $p$ is not a prime number, which contradicts the given condition.
Thus, $t \neq 3$. Then
$$
\begin{array}{l}
t \equiv \pm 1(\bmod 3) \\
\Rightarrow 2 t^{2}+1 \equiv 2 \times 1+1 \equiv 0(\bmod 3) \\
\Rightarrow 3\left|\left(2 t^{2}+1\right) \Rightarrow 3\right| p^{2} \\
\Rightarrow 3 \mid p \Rightarrow p=3 .
\end{array}
$$

In this case, $t=2$, so $x_{0}=3, y_{0}=2^{2}$ satisfies the condition.
In conclusion, $p=3$ is the solution.