Do there exist $2000$ real numbers (not necessarily distinct) such that all of them are not zero and if we put any group containing $1000$ of them as the roots of a monic polynomial of degree $1000$, the coefficients of the resulting polynomial (except the coefficient of $x^{1000}$) be a permutation of the $1000$ remaining numbers? Proposed by Morteza Saghafian
Problem
Source: Iran TST 2012-Third exam-1st day-P2
Tags: algebra, polynomial, pigeonhole principle, algebra proposed
hadikh
15.05.2012 14:50
goodar2006 wrote: Do there exist $2000$ real numbers (not necessarily distinct) such that all of them are not zero and if we put any group containing $1000$ of them as the roots of a monic polynomial of degree $1000$, the coefficients of the resulting polynomial (except the coefficient of $x^{1000}$) be a permutation of the $1000$ remaining numbers?
Let $a_1,a_2,\dots ,a_{2000}$ be real numbers that satisfy the condition and $0<|a_1|\leq |a_2|\leq\dots\leq |a_{2000}|$.
Looking at the coefficient of $x^0$ (in each polynomial), we get product of every $1000$ of them is in $1000$ remaining numbers.
So $|a_{2000}a_{1999}\dots a_{1001}|=|a_i|\leq |a_{1001}|$ since $i\in \{1,2,\dots ,1000\}$, Thus $|a_{2000}a_{1999}\dots a_{1002}|\leq 1$.
And $|a_{1000}a_{999}\dots a_{1}|=|a_i|\geq |a_{1000}|$ since $i\in \{1001,1002,\dots ,2000\}$, Thus $|a_{999}a_{998}\dots a_{1}|\geq 1$.
$\Rightarrow 1\geq |a_{2000}a_{1999}\dots a_{1002}|\geq |a_{999}\dots a_{1}|\geq 1\\ \Rightarrow |a_{2000}a_{1999}\dots a_{1002}|=|a_{999}\dots a_{1}|=1$
From this, it's easy to prove that $|a_i|=1$ for all $1\leq i\leq 2000$. By the pigeonhole principle there exist at least $1000$ of them that are equal, WLOG $a_1=a_2=\dots =a_{1000}$.
Looking at the coefficient of $x^{999}$, we get (sum of every $1000$ of them)$\times (-1)$ is in $1000$ remaining numbers.
But $|a_1+a_2+\dots+a_{1000}|\geq 1000$, contradiction!
Hence the answer is "No"!
Sorry for my poor English!
goodar2006
15.05.2012 15:21
Thanks By not all zero, I meant that it's not possible for all of the $2000$ numbers to be zero. But it can be easily shown that none of the numbers must be zero
Swistak
27.05.2012 22:34
Asserting 1 as an argument in our polynomial we can conclude that $(1-x_1)...(1-x_{1000})=1+x_{1001}+...+x_{2000}$ and similarly $(1-x_{1001})...(1-x_{2000})=1+x_1+...+x_{1000}$, so $(1-x_1)...(1-x_{2000})=(1+x_1+...+x_{1000})(1+x_{1001}+..+x_{2000})=\frac{S^2}{4}-y^2$, where $S=2+x_1+...+x_{2000}$ for some $y$. And of course we can take any other $1000$-element subset of $x_1, ..., x_{2000}$ and sum of it plus $1$ have to be either $\frac{S}{2}-y$ or $\frac{S}{2}+y$, so among $x_1, ..., x_{2000}$ there are $1999$ equal values and ending this solution is obvious.
leader
02.05.2013 20:29
the previous posts give better methods but this one(probably the most obvious one to think of) also works. if among $a_{1},..,a{2000}$ there are $x$ diffrent. If $x>2$ $b_{1}<b_{2}<.....<b_{x}$ then we consider the smallest sum and then sums by changing the biggest element in that sum$b_{r}$ to $b_{r+1},b_{r+2},...b_{x}$ and then $b_{r-1}$ to $b_{r}$ and $b_{r-2}$ to $b_{r-1}$ etc... $b_{1}$ to $b_{2}$ to get $x$ different sums(which is the most there can be). now if $x>r$ then there can only be one $b_{x}$ and $x=r+1$. but then there are $b_{r}$ appears at least $1000$ times and if there are at least $2$ numbers smaller than $b_{r}$ we get at least $x+1$ sums. so we have $1998$ equal to $b_{2}$ but then $b_{1}+b_{3}=2*b_{2}$ and $b_{1}*b_{3}=b_{2}^2$ so $b_{1}=b_{2}=b_{3}$ a contradiction. and if $x=r$ then there are at least $1000$ $b_{x}$ and we easily get at least $x+1$ different sums. When $x<=2$ then $1999$ are equal and it's to end this.