Bounding the Number of Roots of Integer Polynomial

Question

Let $P(x)$ be a non constant polynomial in $\mathbb{Z[x]}$. Let $n$ be the number of roots of $P(x)^2-1 = 0$. Show $n \le \deg P+2$.

Answer 1

Let $m=\deg p$, and let $n$ be the number of roots of $p(x)^2-1$ over $\Bbb Z$. We have by degree considerations that $n\leq 2m$. We want to improve this to $n\leq m+2$. Suppose for contradiction that $n> m+2$.

Suppose the solutions are $\{x_1, x_2, \ldots, x_n\}$. Also suppose that $x_1<x_2<\cdots<x_n$. We then have that $p(x_i)^2=1$ for each of the $x_i$. Thus $p(x_i)=\pm 1$. Let $A$ denote the set of $x_i$ such that $p(x_i)=1$, and let $B$ denote the set of $x_i$ such that $p(x_i)=1$. If $A$ has cardinality $m+1$ or more then $p$ would be equivalently $1$ since $p$ has degree $m$. Also, if $B$ has cardinality $m+1$ or more than $p$ would be equivalently $-1$ for the same reasons. Thus both $A$ and $B$ have at least 3 elements. Suppose $A$ contains $x_1$. If not, we can consider the polynomial $-p(x)$, it will have the same roots and also satisfies $[-p(x)]^2-1$. Then $B$ contains at least 3 elements distinct from $x_1$. In particular there is an $n\in B$ such that $n-x_1\geq 3$. We note that $p(n)-p(x_1)=-1-1=-2$. However, in general if $q(x)$ is an integer polynomial and $a$ and $b$ are two distinct integers we have that $a-b\mid q(a)-q(b)$. Thus we conclude that $n-x_1\mid -2$ which is a contradiction. Thus $n\leq m+2$.