# Two problems about the zeros of a prescribed polynomial

Based on some experiments, I find that the following two statements are correct. But I can not prove this. At the same time, I still can not find the counterexmaples.

Let $p(x)=x^{n}+a_{2}x^{n-2}+a_{3}x^{n-3}+\dots+a_{n-1}x+a_{n}$ be a polynomial with interger coefficients, where $a_{k}\geq0$ for every even $k$ and $a_{k}\leq0$ otherwise. Note here that the coefficient of $x^{n-1}$ is equal to 0. Suppose that there exists some odd number $p$ such that $a_{p}<0$ and $a_{p-1}>0$. Then the following two statements should be true:

1. $p(x)$ has at most one nonzero real root.

2. $p(x)$ has no pure imaginary zeros, i.e. $p(x)$ has no zero in the form $\alpha\textrm{i}$, where $\alpha\neq0$ and $\textrm{i}^{2}=-1$.

I am sorry for losing a condition that $a_{2}\geq5$.

• Any other conditions left out? – Gerry Myerson Mar 1 '11 at 3:34
• @Gerry Myerson,thanks. The conditions needed are all presented. – Shunyi Liu Mar 1 '11 at 3:38

$(x^3-1)(x^3-2)(x^2+a^2)=x^8+a^2x^6-3x^5-3a^2x^3+2x^2+2a^2$ has both two different non-zero real roots and $\pm ai$ as roots.

EDIT polynomial adjusted for adjusted conditions. If $a_2$ needs to be at least $b$, set $a$ so $a^2\geq b$. :)

• @Sándor Kovács, thanks for your counterexmaple! I am sorry that I have lost the condition that $a_{2}\geq5$ in my original problem. So, if this condition is added, then the counterexample you given is not appropriate. – Shunyi Liu Mar 1 '11 at 3:34
• check again. :) – Sándor Kovács Mar 1 '11 at 3:47
• @Sándor Kovács, thank you very much for your elegant counterexample. – Shunyi Liu Mar 1 '11 at 4:03

It is not true that $p(x)$ has at most one nonzero real root. For instance, the polynomial $p(x)=x^5+0.001x^3-100x^2+x-0.001$ has three real roots. For the second claim, the polynomial $p(x)=x^5+x^3-x^2-1$ has roots at $\pm i$.

• He specified that $p(x)$ is a polynomial in integer coefficients, so the first counterexample will not work. – Stanley Yao Xiao Mar 1 '11 at 2:39

With the condition $a_2\ge5$ there's still $x^{16}+5x^{14}+x^{12}+x^8-x^5+x^4-x^3+1$ which vanishes at $i$.

• @Gerry Myerson. Thank you very much for your counterexample. – Shunyi Liu Mar 1 '11 at 3:52