IMO shortlist 1994/A4 solution

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

Answer: 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^hf(x/2), so f(x/2) = x^-hf(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/2f(x) - (x/y)^h/2f(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 A² = 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 A²( (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^hA² = A, 4^kA² = -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