April wrote: 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. Let $ m$ be the smallest element and let $ M$ be the biggest. Then $ \frac{m^2}{M-m} \ge m \iff 2m \ge M$. So if the smallest number is $ k$ the biggest number is $ \le 2k$. Then keep $ M$ as the largest and let $ M_2$ be the second largest. If $ \frac{M_2^2}{M-M_2} = M$, then $ M^2-MM_2 - M_2^2 = 0$ which has no integer solutions. Then $ \frac{M_2^2}{M-M_2} \le M_2 \iff 2M_2 \le M$. So $ 2M_2 \le M \le 2m$ giving $ m = M_2$ and $ M = 2m$. So the only solutions are $ S = \{k,2k\}$ $ \forall k \in \mathbb{N}$

The answer is only $\{x, 2x\}$ for any positive integer $x$. Let $S$ be a set that satisfies the given conditions. Select $x$ to be minimal in $S$, and consider some other element $a \in S$. By minimality of $x$,$$\frac{x^2}{a-x} \geq x \implies a \leq 2x.$$Suppose for the sake of contradiction that there exists an $a < 2x$. Then, the condition implies$$\frac{x^2}{a-x} \leq 2x \implies a \geq \frac{3x}2$$for all $a \in S$. Similarly, because$$\frac{x^2}{a-x} \geq \frac{3x}2$$as all $a \neq 2x \geq \frac 32$, we must have $a \leq \frac{5x}3$. Continuing in this fashion, we may obtain$$\frac{F_{2n} x}{F_{2n-1}} \leq a \leq \frac{F_{2n-1}x}{F_{2n-2}}$$for all positive integers $n$. But $\lim_{n \to \infty} \frac{F_n x}{F_{n-1} x} = \phi$ is irrational, so such a positive integer $a$ cannot exist given a positive integer $x$, as required.

Let $m, M$ denote the smallest and largest elements in the set. Then: $$\frac{m^2}{M-m} \in S \implies \frac{m^2}{M-m} \ge m \implies 2m \ge M.$$Consider the second largest element $m'$, then: $$\frac{m'^2}{M-m'} \in S.$$Either $\frac{m'^2}{M-m'} = M$ or $\frac{m'^2}{M-m'} \le m'$. The first case gives us: $M^2-Mm' - m'^2 = 0$ which implies $M = m' \cdot \frac{1+\sqrt{5}}{2}$ which doesn't have integer solutions. Thus: $$\frac{m'^2}{M-m'} \le m' \implies 2m' \le M.$$Therefore: $2m' \le M \le 2m$ which implies $m'=m$ and $M = 2m$. Therefore, the only solution for $S$ are: $\{n, 2n \}$ where $n \in \mathbb N$.