Let $A$ be a set with at least $3$ integers, and let $M$ be the maximum element in $A$ and $m$ the minimum element in $A$. it is known that there exist a polynomial $P$ such that: $m<P(a)<M$ for all $a$ in $A$. And also $p(m)<p(a)$ for all $a$ in $A-(m,M)$. Prove that $n<6$ and there exist integers $b$ and $c$ such that $p(x)+x^2+bx+c$ is cero in $A$.
Problem
Source: Taiwan 2001
Tags: algebra, polynomial, number theory proposed, number theory
Myth
03.03.2005 21:08
Two questions: what is $A-(m,M)$ and what is $n$?
Pascual2005
04.03.2005 04:44
$n=|A|$, and $A-(m,M)$ is all the elements in $A$ different from $M$ and $m$
Myth
04.03.2005 08:32
Pascual2005 wrote: $n=|A|$, and $A-(m,M)$ is all the elements in $A$ different from $M$ and $m$ I.e. $A-(m,M)=A\setminus\{m,M\}$
Pascual2005
04.03.2005 15:34
thanks! now solve it Myth! I know you will enjoy it!
Myth
04.03.2005 17:22
I guess $P(x)\in \mathbb{Z}[x]$ (otherwise consider $P(x)=x/2$, $m=-100,M=100$). But if $P(x)\in \mathbb{Z}[x]$ then I got complete absurd. We have $|P(M)-P(m)|<M-m$ and on the other hand $M-m\mid P(M)-P(m)$. Thus $P(M)=P(m)$. Therefore we can write $P(x)=P(m)+(x-m)(M-x)Q(x)$, where $Q(x)\in \mathbb{Z}[x]$ and $Q(x)>0$ on $A\setminus\{m,M\}$. Consider $a\in A\setminus\{m,M\}$. Then $M-1\geq P(a)=P(m)+(a-m)(M-a)Q(a)\geq (m+1)+1(M-m-1)\cdot 1=M$ -- contradiction... Thus $A\setminus\{m,M\}=\varnothing$. Maybe there are some typos in signs in your statement? Or my solution is wrong...
Pascual2005
04.03.2005 17:40
Sorry Myth! $P$ has integer coeficients as you said!
dieuhuynh
13.06.2006 11:38
I think MYTH right.have you idea?