9. (16 points) Given the quadratic function
$$
y=f(x)=x^{2}+b x+c \text {. }
$$

the graph passes through the point $(1,13)$, and
$$
y=f\left(x-\frac{1}{2}\right)
$$

is an even function.
(1) Find the analytical expression of $f(x)$.
(2) Does the graph of the function $y=f(x)$ have any points where the x-coordinate is a positive integer and the y-coordinate is a perfect square? If such points exist, find their coordinates; if not, explain why.
(1) Since the function
$$
y=f\left(x-\frac{1}{2}\right)
$$

is an even function, the axis of symmetry of the quadratic function $f(x)$ is
$$
x=-\frac{1}{2} .
$$

Thus, $b=1$.
Given that the graph of the quadratic function $f(x)$ passes through the point $(1,13)$, we have $1+b+c=13$.
Therefore, $c=11$.
Hence, $f(x)=x^{2}+x+11$.
(2) If the graph of the function $y=f(x)$ has a point that meets the requirements, let it be $P\left(m, n^{2}\right)\left(m \in \mathbf{N}_{+}, n \in \mathbf{N}\right)$.
Then $m^{2}+m+11=n^{2}$.
Thus, $4 n^{2}-(2 m+1)^{2}=43$, which means
$$
[2 n+(2 m+1)][2 n-(2 m+1)]=43 \text {. }
$$

Note that 43 is a prime number, and
$$
\begin{array}{l}
2 n+(2 m+1)>2 n-(2 m+1) \text {, } \\
2 n+(2 m+1)>0 \text {. } \\
\text { Then }\left\{\begin{array}{l}
2 n+(2 m+1)=43 \\
2 n-(2 m+1)=1
\end{array}\right. \text {, } \\
\Rightarrow\left\{\begin{array}{l}
m=10, \\
n=11 .
\end{array}\right. \\
\end{array}
$$

Therefore, the graph of the function $y=f(x)$ has a point that meets the requirements, and its coordinates are $(10,121)$.