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Find all polynomials $\sum_{k=0}^na_kx^k$, where $a_k=\pm2$ or $a_k=\pm1$, and $0\leq k\leq n,1\leq n<\infty$, such that they have only real zeroes.

I've been thinking about this question, but I've come to the conclusion that I don't have the requisite math knowledge to actually answer it.

An additional, less-important question. I'm not sure where this problem is from. Can someone find a source?

edit

I'm sorry, I have one more request. If this can be evaluated computationally, can you show me a pen and paper way to do it?

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In Mathematica, for order 4, Last/@Select[{(RootReduce[x/.#]&/@Flatten[Solve[x^Range[0,3].#==0]]),x^Range[0,3‌​].#}&/@Tuples[Drop[Range[-2,2],{3}],{4}],Max[Abs[Sign[Im[First[#]]]]]==0&] gives -2 - 2 x + x^2 + x^3 as the first item. –  Ed Pegg Jan 29 '13 at 21:49
    
{{-2,-2,1,1},{-2,-2,2,1},{-2,-2,2,2},{-2,-1,2,1},{-2,1,2,-1},{-2,2,1,-1},{-2,2,2‌​,-2},{-2,2,2,-1},{-1,-2,1,1},{-1,-2,1,2},{-1,-2,2,1},{-1,-2,2,2},{-1,-1,1,1},{-1,‌​-1,2,1},{-1,-1,2,2},{-1,1,1,-1},{-1,1,2,-2},{-1,1,2,-1},{-1,2,1,-2},{-1,2,1,-1},{‌​-1,2,2,-2},{-1,2,2,-1},{1,-2,-2,1},{1,-2,-2,2},{1,-2,-1,1},{1,-2,-1,2},{1,-1,-2,1‌​},{1,-1,-2,2},{1,-1,-1,1},{1,1,-2,-2},{1,1,-2,-1},{1,1,-1,-1},{1,2,-2,-2},{1,2,-2‌​,-1},{1,2,-1,-2},{1,2,-1,-1},{2,-2,-2,1},{2,-2,-2,2},{2,-2,-1,1},{2,-1,-2,1},{2,1‌​,-2,-1},{2,2,-2,-2},{2,2,-2,-1},{2,2,-1,-1}} are the coefficients, if that helps. –  Ed Pegg Jan 29 '13 at 21:50
    
Computation suggests that there are none if $n>6$. –  Julián Aguirre Jul 4 '13 at 13:29

1 Answer 1

Let $\alpha_1,\alpha_2...\alpha_n$ the real roots. We know:

$$\sum \alpha_i^2=( \sum \alpha_i )^2-2\sum \alpha_i\alpha_j= \left(\frac{a_{n-1}}{a_n}\right)^2-2\left(\frac{a_{n-2}}{a_n}\right)\le 8$$

On the other hand, by AM-GM inequality:

$$\sum \alpha_i^2\ge n \sqrt[n]{|\prod\alpha_i|^2}=n\sqrt[n]{\left|\frac{a_0}{a_n}\right|^2}\ge n\sqrt[n]{\frac{1}{4}}$$

So $8\ge n \sqrt[n]{\frac{1}{4}} \Rightarrow n\le9$. The rest is finite enough.

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If you want to solve the remaining cases by hand, it should be helpful organizing them according to the value of $\left(\frac{a_{n-1}}{a_n}\right)^2-2\left(\frac{a_{n-2}}{a_n}\right)$, so you can get sharper bounds for $n$ in each case. –  Rodrigo Apr 26 at 6:27
    
Very elegant.${}$ –  Antonio Vargas Apr 26 at 21:36

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