To solve the problem, we need to find the intersection points of the curves \( x^2 - y^2 = 4 \) and \( xy = \sqrt{5} \), and then determine the possible number of points \( C \) on the hyperbola \( x^2 - y^2 = 4 \) such that triangle \( ABC \) is equilateral.
### Step 1: Find the intersection points of the curves
1. **Equations of the curves:**
- Curve 1: \( x^2 - y^2 = 4 \)
- Curve 2: \( xy = \sqrt{5} \)
2. **Express \( x \) in terms of \( y \) using the second equation:**
\[
x = \frac{\sqrt{5}}{y}
\]
3. **Substitute \( x \) into the first equation:**
\[
\left(\frac{\sqrt{5}}{y}\right)^2 - y^2 = 4
\]
\[
\frac{5}{y^2} - y^2 = 4
\]
4. **Multiply through by \( y^2 \) to eliminate the fraction:**
\[
5 - y^4 = 4y^2
\]
\[
y^4 + 4y^2 - 5 = 0
\]
5. **Let \( z = y^2 \), then the equation becomes:**
\[
z^2 + 4z - 5 = 0
\]
6. **Solve this quadratic equation using the quadratic formula:**
\[
z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 + 20}}{2} = \frac{-4 \pm \sqrt{36}}{2}
\]
\[
z = \frac{-4 \pm 6}{2}
\]
\[
z = 1 \quad \text{or} \quad z = -5
\]
7. **Since \( z = y^2 \), we discard \( z = -5 \) (not valid) and take \( z = 1 \):**
\[
y^2 = 1 \implies y = \pm 1
\]
8. **Find corresponding \( x \) values:**
- For \( y = 1 \):
\[
x = \frac{\sqrt{5}}{1} = \sqrt{5}
\]
- For \( y = -1 \):
\[
x = \frac{\sqrt{5}}{-1} = -\sqrt{5}
\]
9. **Thus, the intersection points are:**
\[
A(\sqrt{5}, 1) \quad \text{and} \quad B(-\sqrt{5}, -1)
\]
### Step 2: Determine the points \( C \) on the hyperbola such that triangle \( ABC \) is equilateral
1. **The coordinates of points \( A \) and \( B \) are:**
\[
A = (\sqrt{5}, 1), \quad B = (-\sqrt{5}, -1)
\]
2. **The midpoint \( D \) of segment \( AB \):**
\[
D = \left( \frac{\sqrt{5} - \sqrt{5}}{2}, \frac{1 - 1}{2} \right) = (0, 0)
\]
3. **The length of side \( AB \):**
\[
AB = \sqrt{(\sqrt{5} - (-\sqrt{5}))^2 + (1 - (-1))^2} = \sqrt{(2\sqrt{5})^2 + (2)^2} = \sqrt{20 + 4} = \sqrt{24} = 2\sqrt{6}
\]
4. **For triangle \( ABC \) to be equilateral, point \( C \) must be at a distance of \( AB \) from both \( A \) and \( B \).**
5. **The general point \( C \) on the hyperbola \( x^2 - y^2 = 4 \):**
\[
C = (2\sec(\theta), 2\tan(\theta))
\]
6. **To form an equilateral triangle, the slopes of lines \( AC \) and \( BC \) must be such that the angle between them is \( 60^\circ \). This leads to a condition involving the slopes.**
7. **However, since the slopes must satisfy certain conditions and the geometric configuration of points \( A \), \( B \), and \( C \), we find that there are no valid points \( C \) that satisfy these conditions.**
### Conclusion
Thus, the possible number of points \( C \) on the curve \( x^2 - y^2 = 4 \) such that triangle \( ABC \) is equilateral is:
\[
\boxed{0}
\]
