Question is like in title:

For positive integer $n$ I want to construct a polynomial $w(x)$ of degree $n$ with integer coefficients such that there exist $a_1<a_2<\cdots<a_n$ (integer numbers) satisfying $w(a_1)=w(a_2)=\cdots=w(a_n)=1$ and $b_1<b_2<\cdots<b_n$ (also integer numbers) satisfying $w(b_1)=w(b_2)=\cdots=w(b_n)=-1$.

I am not sure if such polynomial even exists for every $n$.

EDIT: $a_i, b_i$ are not predetermined, as someone mentioned in a comment.


This is not possible for $n>3$. In fact, it is not possible to find a degree-$n$ polynomial $w(x)$ which has $n$ different integer roots of $w(x)=1$ and even one integer root (say $b$) of $w(x)=-1$. This is because, if $a_1,...,a_n$ are the roots of $w(x)=1$, $w(x)-1=c(x-a_1)...(x-a_n)$, where $c$ is an integer. Thus $w(b)-1$ is $c$ times a product of $n$ different integers, and since at most two of these can be $\pm 1$, and all others have absolute value at least $2$, if $n>3$ then $w(b)-1$ can't equal $-2$ -- it must have absolute value at least $2^{n-2}$.

[edit] In fact what you asked for is not possible for $n=3$ either. For $w(b)-1=-2$ to have a solution we must have $(b-a_1)(b-a_2)(b-a_3)=-2$ so the three terms are $-1,+1,+2$. But if those are the three terms, then the order of them is determined, so there is at most one solution for $b.$ It is possible for $n=2$: take $w(x)=x^2-3x+1$. Then $w(x)=1$ has two solutions $x=0,3$ and $w(x)=-1$ has two solutions $x=1,2$.

  • $\begingroup$ Great answer, thanks! $\endgroup$
    – Shingle
    Feb 21 '17 at 12:01
  • $\begingroup$ Nice argument (+1). $\endgroup$
    – lulu
    Feb 21 '17 at 12:01

$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$$\newcommand{\Set}[1]{\left\{ #1 \right\}}$Just to complement the excellent answer (+1), if $n = 1$ you have $$ w(x) = C (x - a_{1}) + 1, $$ and $$w(b_{1}) = C (b_{1} - a_{1}) = -2,$$ so there is a solution iff $b_{1} - a_{1} \in \Set{\pm 1, \pm 2}$.

If $n = 2$ you have $$ w(x) = C (x - a_{1})(x - a_{2}) + 1, $$ and need $$ C (b_{1} - a_{1})(b_{1} - a_{2}) = -2 = C (b_{2} - a_{1})(b_{2} - a_{2}), $$ and there are solutions for instance for $$\tag1\label1 b_{1} = a_{1} + 1 = a_{2} - 2, $$ so that $a_{2} = a_{1} + 3$ and $$\tag2\label2 b_{2} = a_{1} + 2 = a_{2} - 1, $$ say $a_{1} = 0, a_{2} = 3, b_{1} = 1, b_{2} = 2$, and then $C = 1$.

Actually, for the case $n = 2$ we have $$ a_{2} - a_{1} = \Size{a_{2} - a_{1}} \le \Size{b_{1} - a_{1}} + \Size{b_{1} - a_{2}} \le 3, $$ so I think the general conditions for the existence of a solution are indeed $\eqref1$ and $\eqref2$.


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