Let $n$ be a positive integer and let $k_0,k_1, \dots,k_{2n}$ be nonzero integers such that $k_0+k_1 +\dots+k_{2n}\neq 0$. Is it always possible to a permutation $(a_0,a_1,\dots,a_{2n})$ of $(k_0,k_1,\dots,k_{2n})$ so that the equation \begin{align*} a_{2n}x^{2n}+a_{2n-1}x^{2n-1}+\dots+a_0=0 \end{align*}has not integer roots?
Problem
Source: All russian olympiad 2016,Day2,grade 11,P5
Tags: number theory, Integer Polynomial, algebra, polynomial
05.05.2016 20:33
Could you please give a link to the original russian text, since as written it can be interpreted as $a_{2n}x^{2n}+a_{2n-1}x^{2n-1}+...+a_0=0$ to have at least one non integer root, which is next to obvious.
05.05.2016 21:44
It should be that the polynomial has no integer roots if I am not mistaken.
06.05.2016 09:56
Polynomial can have some roots, but all roots must be not integer. Here original russian text http://olympiads.mccme.ru/vmo/2016/final/sheets2.pdf
06.05.2016 10:04
Thanks! We will prove there exists a permutation $a_0,a_1,\dots,a_{2n-1}$ of $(k_0,k_1,...,k_{2n-1})$, such that all roots of $a_{2n}x^{2n}+a_{2n-1}x^{2n-1}+...+a_0=0$ are non integers. Arguing by contradiction, suppose for all permutations $(a_0,...a_{2n})$ the corresponding polynomial has at least one integer root. WLOG suppose $|k_{2n}|\geq |k_{i}|\,,\,i=0,1,\dots,2n-1$. Consider the polynomial: \[P(x):=k_{2n}x^{2n}+a_{2n-1}x^{2n-1}+...+a_0\,\,\,\,\,\,\, (*)\]where $a_0,a_1,\dots,a_{2n-1}$ is some permutation of $(k_0,k_1,...,k_{2n-1})$. When $|x|\geq 2$, it holds: \[|k_{2n}x^{2n}|>|a_{2n-1}x^{2n-1}|+...+|a_0|\]It means the only possible integer roots of (*) are $x\in\{-1,0,1\}$. Taking into account the initial assumptions, the only possibility is $x=-1$. Since we can permutate $(k_0,k_1,...,k_{2n-1})$ as we want, it implies $k_0=k_1=\dots=k_{2n-1}$. But then $P(-1)=k_{2n}\neq 0$, a contradiction.
05.07.2020 03:53
It is always possible. Suppose for sake of contradiction that all permutations produced a polynomial with integer roots. Suppose WLOG that $|k_{2n}|\ge|k_0|,\ldots,|k_{2n-1}|$. Then, all permutations with $a_{2n}=k_{2n}$ must have the integer root $x=-1$, as for $|x|\ge 2$, we have \[|k_{2n}|\cdot |x|^{2n} > \sum_{i=0}^{2n-1}|k_i|\cdot |x|^{i},\]and clearly $x=1$ can't be a root by the hypothesis. Thus, \[k_{2n}+a_{2n-2}+a_{2n-4}+\cdots+a_2+a_0 = a_{2n-1}+a_{2n-3}+\cdots+a_3+a_1\]for all permutations $(a_0,\ldots,a_{2n-1})$ of $(k_0,\ldots,k_{2n-1})$. By swapping $a_i$ and $a_j$ where $i$ is even and $j$ is odd, we get that $a_i=a_j$, so in fact $a_0=\cdots=a_{2n-1}=a$. But this implies $k_{2n}=0$, which is the desired contradiction.
06.07.2020 09:00
use this lemma: Let a be a root of P(x) with \[P(x) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ... + {a_1}x + {a_0}\]then \[a < 1 + \mathop {\max }\limits_{1 \leqslant i \leqslant n} \left| {\frac{{{a_i}}}{{{a_n}}}} \right|\]