IMO 1990 SL 25

Origin: TUR

Problem

Let Q+ be the set of positive rational numbers. Construct a function f : Q+ \toQ+ such that f(xf(y)) = f(x) y , for all x, y in Q+.

Solution

Plugging in x = 1 we get f(f(y)) = f(1)/y and hence f(y1) = f(y2) implies y1 = y2 i.e. that the function is bijective. Plugging in y = 1 gives us f(xf(1)) = f(x) ⇒xf(1) = x ⇒f(1) = 1. Hence f(f(y)) = 1/y. Plugging in y = f(z) implies 1/f(z) = f(1/z). Finally setting y = f(1/t) into the original equation gives us f(xt) = f(x)/f(1/t) = f(x)f(t). Conversely, any functional equation on Q+ satisfying (i) f(xt) = f(x)f(t) and (ii) f(f(x)) = x for all x, t \inQ+ also satisfies the original func- tional equation: f(xf(y)) = f(x)f(f(y)) = f(x) y . Hence it suffices to find a function satisfying (i) and (ii). We note that all elements q \inQ+ are of the form q = $n i=1 pai i where pi are prime and ai \inZ. The criterion (a) implies f(q) = f($n i=1 pai i ) = $n i=1 f(pi)ai. Thus it is sufficient to define the function on all primes. For the function to satisfy (b) it is necessary and sufficient for it to satisfy f(f(p)) = 1 p for all primes p. Let qi denote the i-th smallest prime. We define our function f as follows: f(q2k−1) = q2k, f(q2k) = q2k−1 , k \inN . Such a function clearly satisfies (b) and along with the additional condition f(xt) = f(x)f(t) it is well defined for all elements of Q+ and it satisfies the original functional equation.