4. In the Cartesian coordinate system, there is a parabola $y=$ $x^{2}-(5 c-3) x-c$ and three points $A\left(-\frac{1}{2} c, \frac{5}{2} c\right)$, $B\left(\frac{1}{2} c, \frac{9}{2} c\right)$, $C(2 c, 0)$, where $c>0$. There exists a point $P$ on the parabola such that the quadrilateral with vertices $A$, $B$, $C$, and $P$ is a parallelogram. Then the number of such points $P$ is $\qquad$.
4.3.
(1) If $A B$ is the diagonal, then $P_{1}(-2 c, 7 c)$. For $P_{1}$ to be on the parabola, it must satisfy
$$
7 c=(-2 c)^{2}-(5 c-3)(-2 c)-c \text {. }
$$

Solving this, we get $c_{1}=0$ (discard), $c_{2}=1$.
Thus, $c=1$, and at this point, $P_{1}(-2,7)$.
(2) If $B C$ is the diagonal, then $P_{2}(3 c, 2 c)$. Similarly, we get $c=1$, and at this point, $P_{2}(3,2)$.
(3) If $A C$ is the diagonal, then $P_{3}(c,-2 c)$, we get $c=1$, and at this point, $P_{3}(1,-2)$.

In summary, there exist three points $P_{1}(-2,7)$, $P_{2}(3,2)$, and $P_{3}(1,-2)$, such that the quadrilateral with vertices $A$, $B$, $C$, and $P$ is a parallelogram.