IMO 1977 LL POL31

Origin: POL

Problem

Let f be a function defined on the set of pairs of nonzero rational numbers whose values are positive real numbers. Suppose that f satisfies the following conditions: (1) f(ab, c) = f(a, c)f(b, c), f(c, ab) = f(c, a)f(c, b); (2) f(a, 1 −a) = 1. Prove that f(a, a) = f(a, −a) = 1, f(a, b)f(b, a) = 1.

Solution

We obtain from (1) that f(1, c) = f(1, c)f(1, c); hence f(1, c) = 1 and con- sequently f(−1, c)f(−1, c) = f(1, c) = 1, i.e. f(−1, c) = 1. Analogously, f(c, 1) = f(c, −1) = 1. Clearly f(1, 1) = f(−1, 1) = f(1, −1) = 1. Now let us assume that a ̸= 1. Observe that f(x−1, y) = f(x, y−1) = f(x, y)−1. Thus by (1) and (2) we get 1 = f(a, 1 −a)f(1/a, 1 −1/a) = f(a, 1 −a)f  a, 1 −1/a  = f  a, 1 −a 1 −1/a  = f(a, −a).

We now have f(a, a) = f(a, −1)f(a, −a) = 1 \cdot 1 = 1 and 1 = f(ab, ab) = f(a, ab)f(b, ab) = f(a, a)f(a, b)f(b, a)f(b, b) = f(a, b)f(b, a).