Problem 5. Find the number of real solution of the system

$$
\left\lvert\, \begin{aligned}
& x+y+z=3 x y \\
& x^{2}+y^{2}+z^{2}=3 x z \\
& x^{3}+y^{3}+z^{3}=3 y z
\end{aligned}\right.
$$

Sava Grozdev, Svetlozar Doychev
5. Note first that the the triple $(0,0,0)$ is a solution of the system.

If $y=0$, then it follows from the first equation that $x=-z$ and the second one gives that $x=z=0$.

If $y \neq 0$, set $a=\frac{x}{y}$ and $b=\frac{z}{y}$. Then the system becomes

$$
\left\lvert\, \begin{aligned}
& 1+a+b=3 a y \\
& 1+a^{2}+b^{2}=3 a b \\
& y\left(1+a^{3}+b^{3}\right)=3 b
\end{aligned}\right.
$$

Hence $y=\frac{1+a+b}{3 a}$ and therefore

$$
\left\lvert\, \begin{aligned}
& (1+a+b)\left(1+a^{3}+b^{3}\right)=9 a b \\
& 1+a^{2}+b^{2}=3 a b
\end{aligned}\right.
$$

Set $u=a+b$ and $v=a b$. Then

$$
\left\lvert\, \begin{aligned}
& (1+u)\left(1+u^{3}-3 u v\right)=9 v \\
& 1+u^{2}-2 v=3 v
\end{aligned}\right.
$$

Hence $v=\frac{u^{2}+1}{5}$ and plugging it into the first equation gives

$$
0=u^{4}+u^{3}-6 u^{2}+u-2=(u-2)\left(u^{3}+3 u^{2}+1\right)
$$

The case $u=2$ leads to $v=1, a=b=1$ and gives the solution $(x, y, z)=$ $(1,1,1)$. The function $f(u)=u^{3}+3 u^{2}+1$ has a local maximum at $u=-2$ and a local minimum at $u=0$. Since $f(0)=1>0$, the equation $f(u)=0$ has only one real root $u_{0}$ and $u_{0}0$ which shows that the system

$$
\left\lvert\, \begin{aligned}
& a+b=u_{0} \\
& a b=\frac{u_{0}^{2}+1}{5}
\end{aligned}\right.
$$

has two solution.

Thus the given system has four real solutions.