42nd Putnam 1981

p(x) is a real polynomial with at least n distinct real roots greater than 1. [To be precise we can find at least n distinct values a_i > 1 such that p(a_i) = 0. It is possible that one or more of the a_i is a multiple root, and it is possible that there are other roots.] Put q(x) = (x² + 1) p(x) p'(x) + x p(x)² + x p'(x)². Must q(x) have at least 2n - 1 distinct real roots?

Solution

Answer: yes.

Notice that q(x) = (x p(x) + p'(x) )( p(x) + x p'(x) ). The second factor is (x p(x) )'. Now x p(x) has at least n + 1 distinct roots (those of p(x) plus the root x = 0), so its derivative has at least n distinct zeros. To be precise assume that p(x) has exactly m ≥ n values a_i> 1 at which it is zero: 1 < a₁ < a₂ < ... < a_m. Then (x p(x) )' has at least m zeros, one in the interval (0, a₁) and one in each of the intervals (a_i, a_i+1).

We would like to make a similar argument for x p(x) + p'(x). This needs a trick. p(x) exp(x²/2) also has zeros at a_i. Hence its derivative, exp(x₂/2) ( x p(x) + p'(x) ) has at least m-1 roots, one in each interval (a_i, a_i+1). But exp(x²/2) is positive for all real x, so (x p(x) + p'(x) ) has a root in each interval (a_i, a_i+1).

So we are done unless a root k of (x p(x) )' is also a root of (x p(x) + p'(x) ). But in this case k belongs to some open interval (a_i, a_i+1) and we have k p(k) + p'(k) = k p'(k) + p(k). Hence (k - 1)( p(k) - p'(k) ) = 0. But k > 1, so p'(k) = p(k). Hence kp(k) + p(k) = 0, so p(k) = 0. But this contradicts the assumption that the a_i are the only roots greater than 1. Hence there are at least 2m - 1 (which is ≥ 2n - 1) roots.

42nd Putnam 1981

© John Scholes
jscholes@kalva.demon.co.uk
16 Jan 2001