The 1996 Canadian Mathematical Olympiad, Problem 2 is: Find all real solutions to the system of equations
$$
\left\{\begin{array}{l}
\frac{4 x^{2}}{1+4 x^{2}}=y, \\
\frac{4 y^{2}}{1+4 y^{2}}=z, \\
\frac{4 z^{2}}{1+4 z^{2}}=x
\end{array}\right.
$$

The 1996, Issue 5 provided one solution method. Below, we present two clever solutions.
Solution 1: From the problem, we know $x \geqslant 0, y \geqslant 0, z \geqslant 0$. Obviously, $x=0, y=0, z=0$ is one set of solutions to the system of equations.
Assume $x, y, z > 0$.
(1) $\times$ (2) $\times$ (3) gives
$$
\frac{64 x y z}{\left(1+4 x^{2}\right)\left(1+4 y^{2}\right)\left(1+4 z^{2}\right)}=1.
$$

By the AM-GM inequality, we have
$$
1+4 x^{2} \geqslant 4 x, 1+4 y^{2} \geqslant 4 y, 1+4 z^{2} \geqslant 4 z,
$$

where equality holds if and only if $x=y=z=\frac{1}{2}$.
$$
\therefore \frac{64 x y z}{\left(1+4 x^{2}\right)\left(1+4 y^{2}\right)\left(1+4 z^{2}\right)} \leqslant 1,
$$

which is achieved if and only if $x=y=z=\frac{1}{2}$. Therefore, the solutions are:
$$
\begin{array}{l}
(x, y, z)=(0,0,0), \\
(x, y, z)=\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right).
\end{array}
$$

Solution 2: From the problem, we have $x \geqslant 0, y \geqslant 0, z \geqslant 0$. Clearly, $x=0, y=0, z=0$ is one set of solutions to the system of equations.
When $x>0, y>0, z>0$, (1) + (2) + (3) gives
$$
\begin{array}{l}
x+y+z \\
=\frac{4 x^{2}}{1+4 x^{2}}+\frac{4 y^{2}}{1+4 y^{2}}+\frac{4 z^{2}}{1+4 z^{2}} \\
\leqslant \frac{4 x^{2}}{4 x}+\frac{4 y^{2}}{4 y}+\frac{4 z^{2}}{4 z} \text{ (by the AM-GM inequality) } \\
=x+y+z.
\end{array}
$$

The above inequality holds with equality if and only if $x=y=z=\frac{1}{2}$.

Upon verification, $x=\frac{1}{2}, y=\frac{1}{2}, z=\frac{1}{2}$ is another set of solutions.
Therefore, the system of equations has only two sets of solutions.