8.2 Sergey arranged several (more than two) pairwise distinct real numbers in a circle so that each number turned out to be equal to the product of its neighbors. How many numbers could Sergey have arranged?
Answer: 6.

Solution: Let's denote the two adjacent numbers as $a$ and $b$. Then, next to them stands $b / a$, followed by $1 / a, 1 / b, a / b$, and again $a$. Thus, it is impossible to arrange more than 6 numbers.

If 3 numbers can be arranged, then $a=1 / a$, which means $a$ is 1 or -1. In the first case, $b$ and $b / a$ coincide. In the second case, we get the triplet: $-1, b, -b$, where $-1=b * (-b)$, meaning this triplet is $-1, 1, -1$.

If 4 numbers can be arranged, then $a=1 / b$, resulting in the quartet of numbers $a, 1 / a, 1 / a^2, 1 / a$. There are repeating numbers.

If 5 numbers can be arranged, then $a=a / b$, which means $b=1$. This results in the quintet of numbers $a, 1, 1 / a, 1 / a, 1$. There are repeating numbers.

Finally, 6 numbers can be arranged: $2, 3, 3 / 2, 1 / 2, 1 / 3, 2 / 3$.

Criteria: Proving that there are no more than 6 numbers - 3 points.

For considering each case - 1 point each.