IMO 1979 SL 3

Origin: BUL

Problem

Find all polynomials f(x) with real coefficients for which f(x)f(2x2) = f(2x3 + x).

Solution

An obvious solution is f(x) = 0. We now look for nonzero solutions. We note that plugging in x = 0 we get f(0)2 = f(0); hence f(0) = 0 or f(0) = 1. If f(0) = 0, then f is of the form f(x) = xkg(x), where g(0) ̸= 0. Plugging this formula into f(x)f(2x2) = f(2x3 + x) we get 2kx2kg(x)g(2x2) = (2x2 + 1)kg(2x3 + x). Plugging in x = 0 gives us g(0) = 0, which is a contradiction. Hence f(0) = 1. For an arbitrary root \alpha of the polynomial f, 2\alpha3 + \alpha must also be a root. Let \alpha be a root of the largest modulus. If |\alpha| > 1 then |2\alpha3 + \alpha| > 2|\alpha|3 −|\alpha| > |\alpha|, which is impossible. It follows that |\alpha| \leq1 and hence all roots of f have modules less than or equal to 1. But the product of all roots of f is |f(0)| = 1, which implies that all the roots have modulus

1. Consequently, for a root \alpha it holds that |\alpha| = |2\alpha3 −\alpha| = 1. This is possible only if \alpha = \pmı. Since the coefficients of f are real it follows that f must be of the form f(x) = (x2 + 1)k where k \inN0. These polynomials satisfy the original formula. Hence, the solutions for f are f(x) = 0 and f(x) = (x2 + 1)k, k \inN0.