Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


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?

share|cite|improve this question
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

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.

share|cite|improve this answer
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 '14 at 6:27
Very elegant.${}$ – Antonio Vargas Apr 26 '14 at 21:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.