Hi guys,
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$\displaystyle \pm 1$ is rational the rational roots theorem should be of some help
Rational root theorem - Wikipedia, the free encyclopedia
Hello, arsenicbear!
Determine all polynomials that have only real roots and all coefficients are $\displaystyle \pm1$.
I started scribbling and came up a surprising answer.
. . (Or am I totally wrong?)
There are two linear polynomials: .$\displaystyle \begin{array}{ccc}x + 1 \\ x - 1 \end{array}$
There are two cubic polynomials: .$\displaystyle \begin{array}{ccc}x^3-x^2 - x + 1 \\ x^3 + x^2 - x - 1 \end{array}$
I did not include $\displaystyle x^2-1$
It has a zero-coefficient: .$\displaystyle x^2 + {\bf 0}x - 1$
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 $\displaystyle x^n + a_1 x^{n-1} + \ldots + a_n$ is $\displaystyle a_1^2 - 2a_2$ . The product of the squares of the roots is $\displaystyle a_n^2 $. Using the AGM inequality we have $\displaystyle \frac{a_1^2 - 2a_2}n \geqslant \sqrt[n]{a_n^2}.$
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.