## 2. Solve the system of equations

$$
\left\{\begin{array}{l}
x^{2}+1=2 y \\
y^{2}+1=2 z \\
z^{2}+1=2 x
\end{array}\right.
$$
Solution. By collecting the equations of the system, we get

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

The sum of the squares of three numbers is equal to zero if and only if each of the addends is equal to zero, i.e.,

$$
x-1=0, y-1=0 \text { and } z-1=0 \text {. }
$$

Therefore, it follows that $x=1, y=1, z=1$.