IMC 2003 · Day 1 · P4

Clearly $a$ and $b$ must be different since $A$ and $B$ are disjoint.

Let $\{a, b\}$ be a solution and consider the sets $A$, $B$ such that $a \cdot A = b \cdot B$. Denoting $d = (a, b)$ the greatest common divisor of $a$ and $b$, we have $a = d \cdot a_1$, $b = d \cdot b_1$, $(a_1, b_1) = 1$ and $a_1 \cdot A = b_1 \cdot B$. Thus $\{a_1, b_1\}$ is a solution and it is enough to determine the solutions $\{a, b\}$ with $(a, b) = 1$.

If $1 \in A$ then $a \in a \cdot A = b \cdot B$, thus $b$ must be a divisor of $a$. Similarly, if $1 \in B$, then $a$ is a divisor of $b$. Therefore, in all solutions, one of numbers $a$, $b$ is a divisor of the other one.

Now we prove that if $n \ge 2$, then $(1, n)$ is a solution. For each positive integer $k$, let $f(k)$ be the largest non-negative integer for which $n^{f(k)} \mid k$. Then let $A = \{ k : f(k) \text{ is odd} \}$ and $B = \{ k : f(k) \text{ is even} \}$. This is a decomposition of all positive integers such that $A = n \cdot B$.