Find all real-valued functions f on the reals which have at most finitely many zeros and satisfy f(x4 + y) = x3f(x) + f(f(y)) for all x, y.
Solution
Putting x = 0, we get f(y) = f(f(y)) for all y (*).
Putting x = 1, y = 0, we get f(0) = 0. Hence putting y = 0, f(x4) = x3f(x) (**).
If f(1) = 0, then f(2) = f(1 + 1) = f(1) + f(1) = 0, so f(3) = f(1 + 2) = f(1) + f(2) = 0 and so on. Contradiction (only finitely many zeros). So f(1) = k for some non-zero k. By (*), f(k) = k.
Suppose f(h) = 0 for some h not 0 or 1. Then f(h4) = h3f(h) = 0, so f(x) = 0 for any of the distinct values x = h, h4, h16, h64, ... . Contradiction (only finitely many zeros). So f(h) is not 0 for any non-zero h.
Given any x, put z = f(x4) - x4. Then f(x4) = f(f(x4)) = f(x4 + z) = x3f(x) + f(z). But f(x4) = x3f(x), so f(z) = 0. Hence z = 0. So for any positive x we have f(x) = x. But f(x4) = f( (-x)4 ) = - x3 f(-x). Hence f(-x) = - f(x). So f(x) = x for all x.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002