Can we find a sequence of reals a_{i} ≠ 0 such that each polynomial p_{n}(x) = a_{0} + a_{1}x + ... + a_{n}x^{n} has all its roots real and distinct?

**Solution**

Answer: yes.

Take a_{0} = -1, a_{1} = 1. We take even coefficients negative and odd coefficients positive. Suppose we have found a_{0}, a_{1}, ... , a_{n}, so that the polynomial p_{n}(x) has roots at x_{1} < x_{2} < ... < x_{n}. Take points y_{0} < x_{1}, y_{1} between x_{1} and x_{2}, y_{2} between x_{2} and x_{3}, ... , y_{n-1} between x_{n-1} and x_{n}, and y_{n} > x_{n}. Then p_{n}(y_{i}) is positive for i odd and negative for i even. If we now take |a_{n+1}| sufficiently small, then p_{n+1}(y_{i}) will also be positive for i odd and negative for i even. Moreover, if we take y_{n+1} to be sufficiently large, then p_{n+1}(y_{n+1}) will have the opposite sign to p_{n+1}(y_{n}). So p_{n+1}(y_{0}), p_{n+1}(y_{1}), ... , p_{n+1}(y_{n+1}) alternate in sign. Hence p_{n+1}(x) has at least n+1 distinct roots, namely one between y_{i} and y_{i+1} for each of i = 0, 1, 2, ... , n.

© John Scholes

jscholes@kalva.demon.co.uk

1 Jan 2001