Solve the following system of equations:

$$
\begin{aligned}
& x=\frac{1}{2}\left(y+\frac{1}{y}\right) \\
& y=\frac{1}{2}\left(z+\frac{1}{z}\right) \\
& z=\frac{1}{2}\left(t+\frac{1}{t}\right) \\
& t=\frac{1}{2}\left(x+\frac{1}{x}\right)
\end{aligned}
$$
It is clear that if $(x, y, z, t)$ is a solution to the system of equations, then none of them are zero, $x, y, z,$ and $t$ have the same sign, and $(-x, -y, -z, -t)$ is also a solution. Therefore, it is sufficient to find the positive solutions.

The sum of a positive number and its reciprocal is at least 2. Thus, the right-hand side of the equations is at least 1, which means that if $(x, y, z, t)$ is a solution consisting of positive numbers, then each of them is at least 1.

By adding the four equations, we get

$$
x+y+z+t=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}
$$

On the left side, we have the sum of four numbers, each of which is at least 1, and on the right side, we have the sum of their reciprocals, which are four numbers, each of which is at most 1. These can only be equal if $x=y=z=t=1$.

The obtained quadruple is indeed a solution, so the system of equations has exactly two solutions:

$$
x=y=z=t=1 \quad \text { and } \quad x=y=z=t=-1 \text {. }
$$

Imre Bangha (Győr, Czuczor G. Bencés Gymnasium, 2nd grade)