35th IMO 1994 shortlist Problem A4 h and k are reals. Find all real-valued functions f defined on the positive reals such that f(x) f(y) = y h f(x/2) + x k f(y/2) for all x, y. Solution
: f(x) = 0 is always a solution. If h = k, then there is an additional solution f(x) = 2(x/2) h . Consider first the case h = k. Putting x = y, f(x) 2 = 2x h f(x/2), so f(x/2) = x -h f(x) 2 /2. Hence f(x) f(y) = (y/x) h f(x) 2 /2 + (x/y) h f(y) 2 or ( (y/x) h/2 f(x) - (x/y) h/2 f(y) ) 2 = 0. So f(x)/x h = f(y)/y h . This is true for all x, y, so we have f(x) = A x h for some constant A. Substituting back into the original equation gives A 2 = A/2 h + A/2 h , so A = 0 or A = 2 1-h . Checking, it is clear that A = 0, corresponding to f(x) = 0, is a solution. If A = 2 1-h , then f(x) = 2 (x/2) h . The lhs of the original equation then becomes 4 (x/2) h (y/2) h , and the rhs becomes y h 2 (x/4) h + x h 2 (y/4) h = 4 (x/2) h (y/2) h = lhs. So it is indeed a solution. Now consider h and k unequal. We have f(x) f(y) = y h f(x/2) + x k f(y/2) and f(y) f(x) = x h f(y/2) + y k f(x/2). Subtracting, 0 = (y h - y k ) f(x/2) - (x h - x k ) f(y/2). So f(x/2) = A (x h - x k ) for some constant A for all x except possibly x = 1. Substituting back in the original equation, we get A 2 ( (2x) h - (2x) k )( (2y) h - (2y) k ) = A(x h - x k ) y h + A(y h - y k ) x k , except possibly for x or y = 1/2 or 1. Comparing coefficients, 4 h A 2 = A, 4 k A 2 = -A. So A = 0, and hence f(x) = 0 for all x except possibly 1/2 and 1. So, in particular f(2) = 0. Putting x = y = 2 in the original equation gives: 0 = (2 h + 2 k ) f(1), so f(1) = 0 also. Similarly, putting x = y = 1 in the original equation gives f(1/2) = 0. So f(x) = 0 for all x. 35th IMO shortlist 1994 © John Scholes jscholes@kalva.demon.co.uk 18 Sep 2002