
Problem 1. Solve the system of equations in real numbers:

$$
\left\{\begin{array}{l}
(x-1)(y-1)(z-1)=x y z-1 \\
(x-2)(y-2)(z-2)=x y z-2
\end{array}\right.
$$

(Vladimir Bragin)

Answer: $x=1, y=1, z=1$.


Solution 1. By expanding the parentheses and reducing common terms we obtain

$$
\left\{\begin{array}{l}
-(x y+y z+z x)+(x+y+z)=0 \\
-2(x y+y z+z x)+4(x+y+z)=6
\end{array}\right.
$$

From the first equation we can conclude that $x y+y z+z x=x+y+z$. By substituting this into the second equation, we obtain that $x+y+z=3$. We now have to solve the system

$$
\left\{\begin{array}{l}
x+y+z=3 \\
x y+y z+z x=3
\end{array}\right.
$$

If we square the first equation, we get $x^{2}+y^{2}+z^{2}+2(x y+y z+z x)=9$. Hence $x^{2}+y^{2}+z^{2}=3=x y+y z+z x$.

We will prove that if $x^{2}+y^{2}+z^{2}=x y+y z+z x$, than $x=y=z$ :

$$
\begin{gathered}
x^{2}+y^{2}+z^{2}=x y+y z+z x \quad \Longleftrightarrow \\
2 x^{2}+2 y^{2}+2 z^{2}=2 x y+2 y z+2 z x \quad \Longleftrightarrow \\
x^{2}-2 x y+y^{2}+x^{2}-2 x z+z^{2}+y^{2}-2 y z+z^{2}=0 \\
(x-y)^{2}+(x-z)^{2}+(y-z)^{2}=0
\end{gathered}
$$

The sum of three squares is 0 , so all of them are zeroes, which implies $x=y=z$. That means $x=y=z=1$.

Solution 1'. We will show one more way to solve the system (1). Express $z=3-x-y$ from first equation and substitute it into the second one:

$$
\begin{array}{cc}
x y+(y+x)(3-x-y)=3 \quad \Longleftrightarrow \\
x y+3 x+3 y-2 x y-x^{2}-y^{2}=3 & \Longleftrightarrow \\
x^{2}+y^{2}+x y-3 x-3 y+3=0 & \Longleftrightarrow \\
x^{2}+x(y-3)+y^{2}-3 y+3=0
\end{array}
$$

Let us solve it as a quadratic equation over variable $x$ :

$$
\begin{aligned}
x & =\frac{(3-y) \pm \sqrt{(y-3)^{2}-4\left(y^{2}-3 y+3\right)}}{2}= \\
& =\frac{(3-y) \pm \sqrt{y^{2}-6 y+9-4 y^{2}+12 y-12}}{2}= \\
& =\frac{(3-y) \pm \sqrt{-3 y^{2}+6 y-3}}{2}= \\
& =\frac{(3-y) \pm \sqrt{-3(y-1)^{2}}}{2}
\end{aligned}
$$

We can conclude that $y=1$, because otherwise the square root wouldn't exist. It follows that $x=\frac{3-1 \pm 0}{2}=1$, and then $z=1$.
