40th Putnam 1979 Problem A2 R is the reals. For what real k can we find a continuous function f : R → R such that f(f(x)) = k x 9 for all x.
Evidently k 1/4 x 3 works for any k ≥ 0. For k ≠ 0, f must be (1, 1) [otherwise we would get x 9 = y 9 for x ≠ y, contradiction.] But f is continuous, so it is strictly monotonic. If it is strictly monotonic increasing, then so is f(f(x)). But if it is strictly monotonic decreasing, then f(f(x)) is strictly monotonic increasing . So, either way, f(f(x)) is strictly monotonic increasing. If k < 0, then k x 9 is strictly monotonic decreasing, so we cannot have k < 0. 40th Putnam 1979 © John Scholes jscholes@kalva.demon.co.uk 4 Dec 1999