4th Irish 1991 Problem A5 Find all polynomials x n + a 1 x n-1 + ... + a n such that each of the coefficients a 1 , a 2 , ... , a n is ±1 and all the roots are real.
1, -1, x+1, x-1, x 2 +x-1, x 2 -x-1, x 3 +x 2 -x-1, x 3 -x 2 -x+1 Solution Let the roots be r 1 , r 2 , ... , r n . Then ∑ r i 2 = (∑ r i ) 2 - 2 ∑ r i r j = a 1 2 - 2a 2 = 1 + 2 or 1 - 2. But ∑ r i 2 ≥ 0, so it must be 3. Also ∏ r i 2 = (∏ r i ) 2 = a n 2 = 1. Since all r i 2 are non-negative, AM/GM applies. If n > 3 it would be violated, so n = 0, 1, 2, or 3. The cases n = 0, 1 are trivial. If n = 2 there are 4 possible equations: x 2 +x+1 roots -1/2 ± i(√3)/2, x 2 -x+1 roots 1/2 ± i(√3)/2, x 2 +x-1 roots -1/2 ±(√5)/2, x 2 -x-1 roots 1/2 ±(√5)/2. For n=3 AM/GM is violated unless all r i 2 are equal and hence all equal 1. So each r i is ±1. (1,1,1) and (-1,-1,-1) do not work. (1,1,-1) gives (x-1) 2 (x+1) = (x-1)(x 2 -1) = x 3 -x 2 -x+1. Similarly, (1,-1,-1) gives (x+1) 2 (x-1) = (x+1)(x 2 -1) = x 3 +x 2 -x-1. 4th Irish 1991 © John Scholes jscholes@kalva.demon.co.uk 6 February 2004 Last updated/corrected 6 Feb 04