### IMO 1986

Problem B2

Find all functions f defined on the non-negative reals and taking non-negative real values such that: f(2) = 0, f(x) ≠ 0 for 0 ≤ x < 2, and f(xf(y)) f(y) = f(x + y) for all x, y.

Solution

f(x+2) = f(xf(2)) f(2) = 0. So f(x) = 0 for all x ≥ 2.

f(y) f((2-y)f(y)) = f(2) = 0. So if y < 2, then f((2-y) f(y)) = 0 and hence (2 - y) f(y) ≥ 2, or f(y) ≥ 2/(2 - y).

Suppose that for some y0 we have f(y0) > 2/(2 - y0), then we can find y1 > y0 (and y1 < 2) so that f(y0) = 2/(2 - y1). Now let x1 = 2 - y1. Then f(x1f(y0)) = f(2) = 0, so f(x1 + y0) = 0. But x1 + y0 < 2. Contradiction. So we must have f(x) = 2/(2 - x) for all x < 2.

We have thus established that if a function f meets the conditions then it must be defined as above. It remains to prove that with this definition f does meet the conditions. Clearly f(2) = 0 and f(x) is non-zero for 0 ≤ x < 2. f(xf(y)) = f(2x/(2 - y)). If 2x/(2 - y) ≥ 2, then f(xf(y)) = 0. But it also follows that x + y ≥ 2, and so f(x + y) = 0 and hence f(xf(y)) f(y) = f(x + y) as required. If 2x/(2 - y) < 2, then f(xf(y)) f(y) = 2/(2 - 2x/(2-y)) 2/(2 - y) = 2/(2 - x - y) = f(x + y). So the unique function satisfying the conditions is:

f(x) = 0 for x ≥ 2, and 2/(2 - x) for 0 ≤ x < 2.

Solutions are also available in   István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.