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 x9 for all x.



Evidently k1/4 x3 works for any k ≥ 0.

For k ≠ 0, f must be (1, 1) [otherwise we would get x9 = y9 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 x9 is strictly monotonic decreasing, so we cannot have k < 0.



40th Putnam 1979

© John Scholes
4 Dec 1999