IMO 1997 SL 11

Origin: NET

Problem

Let P(x) be a polynomial with real coefficients such that P(x) > 0 for all x \geq0. Prove that there exists a positive integer n such that (1 + x)nP(x) is a polynomial with nonnegative coefficients.

Solution

All real roots of P(x) (if any) are negative: say −a1, −a2, . . . , −ak. Then P(x) can be factored as P(x) = C(x + a1) \cdot \cdot \cdot (x + ak)(x2 −b1x + c1) \cdot \cdot \cdot (x2 −bmx + cm), (1) where x2 −bix + ci are quadratic polynomials without real roots. Since the product of polynomials with positive coefficients is again a poly- nomial with positive coefficients, it will be sufficient to prove the result for each of the factors in (1). The case of x + aj is trivial. It remains only to prove the claim for every polynomial x2 −bx + c with b2 < 4c. From the binomial formula, we have for any n \inN, (1 + x)n(x2 −bx + c) = n+2  i=0  n i −2  −b  n i −1 

• c n i  xi = n+2  i=0 Cixi, where Ci = n!
  (b + c + 1)i2 −((b + 2c)n + (2b + 3c + 1))i + c(n2 + 3n + 2)  xi i!(n −i + 2)! . The coefficients Ci of xi appear in the form of a quadratic polynomial in i depending on n. We claim that for large enough n this polynomial has negative discriminant, and is thus positive for every i. Indeed, this discriminant equals D = ((b + 2c)n + (2b + 3c + 1))2 −4(b + c + 1)c(n2 + 3n + 2) = (b2 −4c)n2 −2Un + V , where U = 2b2 + bc + b −4c and V = (2b + c + 1)2 −4c, and since b2 −4c < 0, for large n it clearly holds that D < 0.