6. Find all sets of numbers $x_{1}, x_{2}, \ldots, x_{n+1}$ such that $x_{1}=x_{n+1}$ and for all $k=1, \ldots, n$ the equality

$$
2 \log _{2} x_{k} \cdot \log _{2} x_{k+1}-\log _{2}^{2} x_{k}=9
$$
Answer: $x_{k}=8, k=1, \ldots, n+1$, or $x_{k}=\frac{1}{8}, k=1, \ldots, n+1$.

Solution. We need to solve the system of equations

$$
\left\{\begin{array}{l}
9-2 \log _{2} x_{1} \cdot \log _{2} x_{2}+\log _{2}^{2} x_{1}=0 \\
9-2 \log _{2} x_{2} \cdot \log _{2} x_{3}+\log _{2}^{2} x_{2}=0 \\
\cdots \\
9-2 \log _{2} x_{n-1} \cdot \log _{2} x_{n}+\log _{2}^{2} x_{n-1}=0 \\
9-2 \log _{2} x_{n} \cdot \log _{2} x_{1}+\log _{2}^{2} x_{n}=0
\end{array}\right.
$$

The substitution $a_{k}=\frac{1}{3} \log _{2} x_{k}, k=1, \ldots, n$, transforms the system into

$$
\left\{\begin{array}{l}
a_{1}+a_{1}^{-1}=2 a_{2} \\
\cdots \\
a_{n-1}+a_{n-1}^{-1}=2 a_{n} \\
a_{n}+a_{n}^{-1}=2 a_{1}
\end{array}\right.
$$

From the inequality for the sum of reciprocals, it follows that either $a_{k} \geqslant 1, k=1, \ldots, n$, or $a_{k} \leqslant-1$, $k=1, \ldots, n$. The solutions to the system are the sets $a_{k}=1, k=1, \ldots, n$ and $a_{k}=-1, k=1, \ldots, n$. Adding all the equations of the system and moving all terms to one side of the equality, we get

$$
a_{1}-a_{1}^{-1}+\cdots+a_{n-1}-a_{n-1}^{-1}+a_{n}-a_{n}^{-1}=0
$$

The left side of the last equality is greater than zero if at least one $a_{k}>1$, and less than zero if at least one $a_{k}<-1$, so the system has no other solutions besides the ones mentioned above.

Answer to variant 2: $x_{k}=9, k=1, \ldots, n+1$, or $x_{k}=\frac{1}{9}, k=1, \ldots, n+1$.

Answer to variant 3: $x_{k}=2, k=1, \ldots, n+1$, or $x_{k}=\frac{1}{2}, k=1, \ldots, n+1$.

Answer to variant 4: $x_{k}=3, k=1, \ldots, n+1$, or $x_{k}=\frac{1}{3}, k=1, \ldots, n+1$.