13.5.17 Xuan Tai Given the hyperbola $C: x^{2}+2 x-m y^{2}=0 (m>0)$ passing through the origin $O$, is there any two points $A$ and $B$ on $C$ such that $\triangle A O B$ is an isosceles right triangle with $O$ as the right-angle vertex? If not, explain the reason; if so, how many different isosceles right triangles are there?
According to the problem, we can set \( l_{OA}: y = kx \), where \( k \) exists and \( k \neq 0 \), then \( l_{OB}: y = -\frac{1}{k}x \). From the system of equations
\[
\left\{
\begin{array}{l}
y = kx, \\
x^2 + 2x - my^2 = 0,
\end{array}
\right.
\]
we get
\[
(1 - mk^2)x^2 + 2x = 0.
\]
Thus, \( 1 - mk^2 \neq 0 \), so \( x_A = \frac{-2}{1 - mk^2} \), and
\[
|OA| = \sqrt{1 + k^2} \left| \frac{2}{1 - mk^2} \right|.
\]
Similarly,
\[
|OB| = \sqrt{1 + \frac{1}{k^2}} \left| \frac{2}{1 - m \frac{1}{k^2}} \right| = \sqrt{1 + k^2} \left| \frac{2k}{k^2 - m} \right|.
\]
Since \( \triangle AOB \) is an isosceles right triangle with \( O \) as the right-angle vertex, we have \( |OA| = |OB| \), so
\[
\sqrt{1 + k^2} \left| \frac{2}{1 - mk^2} \right| = \sqrt{1 + k^2} \left| \frac{2k}{k^2 - m} \right|, \text{ i.e., } \frac{1}{1 - mk^2} = \pm \frac{k}{k^2 - m}.
\]

Assume \( k > 0 \). If \( \frac{1}{1 - mk^2} = \frac{k}{k^2 - m} \), then
\[
(k - 1)\left[mk^2 + (m + 1)k + m\right] = 0.
\]
Since \( mk^2 + (m + 1)k + m > 0 \), we have \( k = 1 \). If \( \frac{1}{1 - mk^2} = -\frac{k}{k^2 - m} \), then
\[
(k + 1)\left[mk^2 - (m + 1)k + m\right] = 0.
\]
Since \( k > 0 \), we have \( mk^2 - (m + 1)k + m = 0 \). By the discriminant \( \Delta = (m + 1)^2 - 4m^2 = -3m^2 + 2m + 1 > 0 \) and \( m > 0 \), we get \( 0 < m < 1 \). The roots of the quadratic equation are
\[
k_1 + k_2 = \frac{m + 1}{m} > 1, \quad k_1 k_2 = 1 > 0,
\]
so \( k_1, k_2 > 0 \) and \( k_1 \neq 1, k_2 \neq 1 \).

In summary, when \( 0 < m < 1 \), there are two isosceles right triangles; when \( m > 1 \), there is one isosceles right triangle; when \( m = 1 \), \( k = 1 \), then \( 1 - mk^2 = 0 \), so there is no such isosceles right triangle.