I am currently stuck on this Putnam problem and was wondering if anyone had some tips/strategies to attack it?
Determine all polynomials that have only real roots and all coefficients are +/-1?
Since is rational the rational roots theorem should be of some help
Rational root theorem - Wikipedia, the free encyclopedia
Determine all polynomials that have only real roots and all coefficients are .
I started scribbling and came up a surprising answer.
. . (Or am I totally wrong?)
There are two linear polynomials: .
There are two cubic polynomials: .
I did not include
It has a zero-coefficient: .
Did I miss any?
This problem has been bugging me. Since I could make no progress on it, I eventually resorted to a Google search for a solution, which I found here.
We may assume that the leading coefficient is +1. The sum of the squares of the roots of is . The product of the squares of the roots is . Using the AGM inequality we have
Since the coefficients are ±1 that inequality is (1 ± 2)/n ≥ 1, hence n ≤ 3.
Clever, huh? No wonder I couldn't do it.
So there are two of these polynomials in each of degree 1, 2, 3, together with their negatives, and no others, giving a total of 12 such polynomials.