Solution to Exercise 6: Let $a_{1}, a_{2}, \ldots, a_{n}$ be these numbers, with the convention that $a_{n+1}=a_{1}, a_{n+2}=$ $a_{2}$, etc. By hypothesis, we have $a_{i} a_{i+2}=a_{i+1}$ and $a_{i+1} a_{i+3}=a_{i+2}$ for all $i$. Multiplying these two equalities, we get $a_{i} a_{i+1} a_{i+2} a_{i+3}=a_{i+1} a_{i+2}$, hence $a_{i} a_{i+3}=1$. Similarly, $a_{i+3} a_{i+6}=1$, so $a_{i}=a_{i+6}$. Since the integers are all distinct, this implies that 6 is a multiple of $n$, so $n=3$ or $n=6$.

If $n=3$, then $a_{i} a_{i+3}=1$ gives that $a_{i}^{2}=1$ for all $i$, so $a_{i}= \pm 1$. Therefore, at least two of these three numbers are equal, either to 1 or to -1. This contradicts the fact that the numbers are all distinct.

If $\mathrm{n}=6$ then it is possible, for example, with the numbers $2,3, \frac{3}{2}, \frac{1}{2}, \frac{1}{3}, \frac{2}{3}$.
Exercise 6. Determine all integers $n \geqslant 3$ such that it is possible to place $n$ distinct real numbers on a circle so that each of these numbers is the product of its two neighbors.