5. In the set of real numbers, solve the system of equations

$$
\begin{aligned}
& x^{2006}+y^{2006}+z^{2006}=2 \\
& x^{2007}+y^{2007}+z^{2007}=2 \\
& x^{2008}+y^{2008}+z^{2008}=2
\end{aligned}
$$
## Solution.

Let's add the first and third equations and subtract the second one multiplied by 2:

$$
\begin{aligned}
& \left(x^{2006} + y^{2006} + x^{2006}\right) + \left(x^{2008} + y^{2008} + z^{2008}\right) - 2 \cdot \left(x^{2007} + y^{2007} + z^{2007}\right) \\
= & 2 + 2 - 2 \cdot 2 \\
\Leftrightarrow & \left(x^{2006} + x^{2008} - 2 \cdot x^{2007}\right) + \left(y^{2006} + y^{2008} - 2 \cdot y^{2007}\right) + \left(z^{2006} + z^{2008} - 2 \cdot z^{2007}\right) = 0 \\
\Leftrightarrow & x^{2006} \cdot (x-1)^2 + y^{2006} \cdot (y-1)^2 + z^{2006} \cdot (z-1)^2 = 0
\end{aligned}
$$

The sum of non-negative expressions is equal to 0, so each of them must be equal to 0. Therefore, \( x, y, z \in \{0, 1\} \).

(15 points)

The final solutions to the system are \((x, y, z) \in \{(0, 1, 1), (1, 0, 1), (1, 1, 0)\}\). (5 points)

Notes.

- For subtracting the first and second equations

$$
x^{2006} + y^{2006} + z^{2006} - x^{2007} - y^{2007} - z^{2007} = 0
$$

the student gets 2 points.

- For correctly identifying all three solutions without providing an argument that there are no other real solutions, 2 points are given.