IMO 1998 SL 15

Origin: AUS

Problem

Determine all pairs (a, b) of real numbers such that a\lfloorbn\rfloor= b\lflooran\rfloor for all positive integers n. (Note that \lfloorx\rfloordenotes the greatest integer less than or equal to x.)

Solution

The condition is obviously satisfied if a = 0 or b = 0 or a = b or a, b are both integers. We claim that these are the only solutions.

Suppose that a, b belong to none of the above categories. The quotient a/b = \lfloora\rfloor/\lfloorb\rflooris a nonzero rational number: let a/b = p/q, where p and q are coprime nonzero integers. Suppose that p ̸\in{−1, 1}. Then p divides \lflooran\rfloorfor all n, so in particular p divides \lfloora\rfloorand thus a = kp + \epsilon for some k \inN and 0 \leq\epsilon < 1. Note that \epsilon ̸= 0, since otherwise b = kq would also be an integer. It follows that there exists an n \inN such that 1 \leqn\epsilon < 2. But then \lfloorna\rfloor= \lfloorknp + n\epsilon\rfloor= knp + 1 is not divisible by p, a contradiction. Similarly, q ̸\in{−1, 1} is not possible. Therefore we must have p, q = \pm1, and since a ̸= b, the only possibility is b = −a. However, this leads to \lfloor−a\rfloor= −\lfloora\rfloor, which is not valid if a is not an integer.