51st Putnam 1990

Problem B5

Can we find a sequence of reals ai ≠ 0 such that each polynomial pn(x) = a0 + a1x + ... + anxn has all its roots real and distinct?



Answer: yes.

Take a0 = -1, a1 = 1. We take even coefficients negative and odd coefficients positive. Suppose we have found a0, a1, ... , an, so that the polynomial pn(x) has roots at x1 < x2 < ... < xn. Take points y0 < x1, y1 between x1 and x2, y2 between x2 and x3, ... , yn-1 between xn-1 and xn, and yn > xn. Then pn(yi) is positive for i odd and negative for i even. If we now take |an+1| sufficiently small, then pn+1(yi) will also be positive for i odd and negative for i even. Moreover, if we take yn+1 to be sufficiently large, then pn+1(yn+1) will have the opposite sign to pn+1(yn). So pn+1(y0), pn+1(y1), ... , pn+1(yn+1) alternate in sign. Hence pn+1(x) has at least n+1 distinct roots, namely one between yi and yi+1 for each of i = 0, 1, 2, ... , n.



51st Putnam 1990

© John Scholes
1 Jan 2001