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?

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- September 3rd 2010, 12:16 PMarsenicbearPutnam Problem
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? - September 3rd 2010, 01:22 PMTheEmptySet
Since is rational the rational roots theorem should be of some help

Rational root theorem - Wikipedia, the free encyclopedia - September 3rd 2010, 07:25 PMSoroban
Hello, arsenicbear!

Quote:

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 didinclude*not*

It has a zero-coefficient: .

Did I miss any?

- September 5th 2010, 01:37 PMOpalg
- September 7th 2010, 08:42 AMebaines
Here are two more quadratics:

- September 8th 2010, 12:01 PMOpalg
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.