Find all finite sets of positive integers with at least two elements such that for any two numbers $a, b(a>b)$ belonging to the set, the number $\frac{b^{2}}{a-b}$ belongs to the set, too.

Answer: $X=\{a, 2a\}$, where $a$ is an arbitrary nonnegative integer.
Let $X$ be a set we seek for, and $a$ be its minimal element. For each other element $b$ we have $\frac{a^{2}}{b-a} \geq a$, hence $b \leq 2 a$. Therefore all the elements of $X$ belong to the interval $[a, 2 a]$. So the quotient of any two elements of $X$ is at most 2.

Now consider two biggest elements $d$ and $c, c<d$. Since $d \leq 2 c$ we conclude that $\frac{c^{2}}{d-c} \geq c$. Hence $\frac{c^{2}}{d-c}=d$ or $\frac{c^{2}}{d-c}=c$. The first case is impossible because we obtain an equality $(c / d)^{2}+(c / d)-1=0$, which implies that $c / d$ is irrational. Therefore we have the second case and $c^{2}=d c-c^{2}$, i.e. $c=d / 2$. Thus the set $X$ could contain only one element except $d$, and this element should be equal to $d / 2$. It is clear that all these sets satisfy the condition of the problem.