3. Find all finite sets $S$ of positive integers with at least two elements, such that if $m>n$ are two elements of $S$, then

$$
\frac{n^{2}}{m-n}
$$

is also an element of $S$.

Answer: $S=\{s, 2 s\}$, for a positive integer $s$.
Solution(Arnaud):

Trying to apply number theoretical methods to deduce something from the fact that $m-n$ divides $n^{2}$ does not seem to lead anywhere. Instead, we will try to find the extreme values that the quotient $n^{2} /(m-n)$ can achieve. This will give us some interesting bounds on the elements of $S$.

(a) First, since $S$ contains at least two elements, we take $l>s$ to be the largest and smallest elements of $S$. Since $s^{2} /(l-s)$ also belongs to $S$ we must have

$$
\frac{s^{2}}{l-s} \geq s \quad \Longrightarrow \quad 2 s \geq l
$$

(b) Now, consider $kk$.

Proof. Assume such a solution exists, then it also solves

$$
l^{2}=k^{2}+k l
$$

Since the equation is homogeneous (of degree 2), we can assume that $\operatorname{gcd}(k, l)=1$. Since we must have $l \mid k^{2}$, we deduce that $l=1$. This is a contradiction since $k$ is a positive integer with $k<l$.

Combining the two relations above, and using that $s \leq k$, we get

$$
l \leq 2 s \leq 2 k \leq l
$$

Therefore $k=s$ and $S$ contains exactly two elements. Moreover, the inequalities above also implies $l=2 s$. So $S=\{s, 2 s\}$ as desired.

Any such set satisfies the desired property because

$$
\frac{s^{2}}{2 s-s}=s \in S
$$