33rd IMO shortlist 1992/2 solution

33rd IMO 1992 shortlist

Problem 2

a and b are positive reals. Let X be the set of non-negative reals. Show that there is a unique function f:X → X such that f( f(x) ) = b(a + b)x - af(x) for all x.

Answer

f(x) = bx

Solution

Put u₀ = k, u_n+1 = f(u_n). Then u_n+2 - b(a+b)u_n+1 + au_n = 0. That is a standard linear recurrence relation, so we can immediately solve it. The associated quadratic factorises as (x - b)(x + a + b) = 0, so the general solution is u_n = Abⁿ + B(-a-b)ⁿ. Hence u_n/(a+b)ⁿ = A(b/(a+b))ⁿ + (-1)ⁿB. But (b/(a+b))ⁿ → 0 and lhs is always non-negative. Hence B = 0. So A = k. Hence f(k) = u₁ = bk. It is easy to check that this does indeed satisfy the relation given.

33rd IMO shortlist 1992