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 y_{0} we have f(y_{0}) > 2/(2 - y_{0}), then we can find y_{1} > y_{0} (and y_{1} < 2) so that f(y_{0}) = 2/(2 - y_{1}). Now let x_{1} = 2 - y_{1}. Then f(x_{1}f(y_{0})) = f(2) = 0, so f(x_{1} + y_{0}) = 0. But x_{1} + y_{0} < 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.

© John Scholes

jscholes@kalva.demon.co.uk

24 Oct 1998