$x\in \mathbb{R} : P(x) \in \mathbb{Z} \Leftrightarrow Q(x) \in \mathbb{Z}$. Prove that $P(x)-Q(x)=c \in \mathbb{Z}$ or...

Question

$P(x),Q(x)$ are two polynomials such that $x\in \mathbb{R} : P(x) \in \mathbb{Z} \Leftrightarrow Q(x) \in \mathbb{Z}$. Prove that $P(x)-Q(x)=c$ or $P(x)+Q(x)=d, $ where $c,d \in \mathbb{Z}$.

Answer 1

Suppose that $P$ and $Q$ are two polynomials of degree greater than $1$.

Then there is some $a$ such that $P(a),Q(a)\in\mathbb Z$, and $P$ and $Q$ are both strictly monotone on $[a,+\infty)$

Suppose that they are both increasing. Otherwise, multiply the decreasing ones by $-1$.

We will now prove that $P-Q$ is bounded on $[a,\infty)$

Let $(u_n)_{n\in\mathbb N}$ be an increasing sequence for the points on $[a,\infty)$ where $P$ attains a value in $\mathbb Z$, and $(v_n)_{n\in\mathbb N}$ be an increasing sequence for the points on $[a,\infty)$ where $Q$ attains a value in $\mathbb Z$.

First, here are some interesting properties about $(u_n)_{n\in\mathbb N}$ and $(v_n)_{n\in\mathbb N}$.

Clearly, $(u_n)_{n\in\mathbb N}=(v_n)_{n\in\mathbb N}$

$(u_n)_{n\in\mathbb N}$ goes to $\infty$, so $[a,\infty)=\displaystyle\bigcup_{n\in \mathbb N}[u_n,u_{n+1})$

$$P(u_{n+1})=P(u_n)+1$$

$$Q(v_{n+1})=P(v_n)+1$$

These two give us $$P(u_{n+1})-Q(u_{n+1}) = P(u_{n})-Q(u_{n})$$

So $P(u_{n})-Q(u_{n})$ is constant. We'll call it $c$.

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Let $x\in [a,\infty)$, then $x\in[u_p,u_{p+1})$, for some $p\in \mathbb N$. $$\begin{align} P(u_p) \leq P(x) \leq P(u_{p+1})& =P(u_p)+1 \\ Q(u_p) \leq Q(x) \leq Q(u_{p+1})& = Q(u_p)+1 \end{align}$$

So now we have: $$\begin{align} P(u_p)-Q(u_p)-1 &\leq P(x)-Q(x) \leq P(u_p)-Q(u_p)+1 \\ c-1 &\leq P(x)-Q(x) \leq c+1 \end{align}$$

So $P-Q$ is bounded on $[a,\infty)$

Therefore, $P-Q$ is constant, and this constant is in $\mathbb Z$, since $(P-Q)(a)\in Z$.

Answer 2

It is not true; Define:
$P(x)=1/3$, and $Q(x)=1/5$ .
then
$P(x) \notin \mathbb{Z}$ and $Q(x) \notin \mathbb{Z}$ but $P(x)-Q(x)=\cdots$ and $P(x)+Q(x)=\cdots$