Find all pairs of real numbers (x, y) such that x [ny] = y [nx] for all positive integers n.
Solution
Answer: the only solutions are: (1) x = y, (2) one of x, y is 0, (3) x and y are both integers.
Obviously the (1), (2), (3) above are solutions. We show there are no others. So suppose x and y are non-zero and unequal. We show by induction that [2nx] = 2n[x] and [2ny] = 2n[y]. Take n = 1. We have 2 [x] ≤ 2x < 2 [x] + 2, so [2x] = 2 [x] or 2 [x] + 1. If [2x] = 2[x] + 1, then x [2y] = y [2x] = 2y [x] + y = 2x [y] + y. So [2y] = 2 [y] + y/x. Hence y = x. Contradiction. So [2x] = 2 [x]. Similarly, [2y] = 2 [y]. So the result is true for n = 1.
Suppose the result is true for n. We have x [2n+1y] = y [2n+1x]. As before, [2n+1x] must be 2 [2nx] or 2 [2nx] + 1. If it is 2 [2nx] + 1, then x [2n+1y] = 2y [2nx] + y = 2n+1y [x] + y = 2n+1x [y] + y, so [2n+1y] = 2n+1 [y] + y/x. Hence y = x. Contradiction. So the result is established for n+1 and hence for all n. But that means that x and y must be integers. So there are no other solutions.
© John Scholes
jscholes@kalva.demon.co.uk
30 Aug 2002