13. $(x, y)=\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)\left(x_{1}<x_{2}\right)$ are two integer solutions of the equation $x^{2}-y^{2}-2 x+6 y-8=0$. In the Cartesian coordinate system, point $A\left(x_{1}, y_{1}\right)$ and point $B\left(x_{2}, y_{2}\right)$ are symmetric about $P(1,3)$, and point $C(5,-1)$. If $\triangle A B C$ has exactly 21 integer points (points with integer coordinates) on its sides and inside, then the coordinates of point $A$ are $\qquad$.
13. $(-1,1)$.

According to the problem, we have
$$
\begin{array}{l}
(x-y+2)(x+y-4)=0 \\
\Rightarrow y=x+2 \text { or } y=-x+4 .
\end{array}
$$

Since point $P(1,3)$ is exactly the intersection of the two lines, and point $C(5,-1)$ lies on the line $y=-x+4$, therefore, points $A$ and $B$ lie on the line $y=x+2$.

There are a total of 5 integer points on line segment $PC$, so by symmetry, point $A(-1,1)$.