28th VMO 1990

The polynomial p(x) with degree at least 1 satisfies p(x) p(2x²) = p(3x³ + x). Show that it does not have any real roots.

Solution

Suppose it has a positive root α, then 3α3+α is another root, which is larger than α. Proceeding, we get infinitely many roots. Contradiction. So there are no positive roots. Similarly, there are no negative roots. So the only possibility is 0. Suppose bx^m is the lowest power of x in p(x). Then the lowest power of x in p(x)p(2x²) is x^3m, but the lowest power of x in p(3x³ + x) is x^m, so m = 0 and hence p(0) = b ≠ 0, so 0 is not a root.

Thanks to Suat Namli

28th VMO 1990

© John Scholes
jscholes@kalva.demon.co.uk
13 February 2004
Last corrected/updated 13 Feb 04