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

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 $(P-Q)(a)\in Z$.
A: 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 $P(x)+Q(x)=\cdots$
