Integer value for a polynomial for values of $x \in \mathbb{R/Z}$ $P(x)$  be a polynomial with real coefficients.It is given that $\text{deg}P \ge 2$.prove that it is not possible that whenever $P(x)$ is an integer, $x$ is an integer or equivalenty there exist $x_0 \in \mathbb{R/Z}$ such that $P(x_0) \in \mathbb{Z}$.
I could make no considerable progress on this problem
 A: Without loss of generality we can suppose that $\lim_{x \to \infty} P(x)= + \infty$.
This assumptions tells us that $\lim_{x \to \infty} P'(x)= + \infty$ as well: that's because the derivative of $P$ is a polynomial of degree at least $1$.
Pick two consecutive integers $n$  and $n+1$ large enough, say with the property that $P'(x) \ge 3$ in the interval $[n,n+1]$. Then by MVT $$|P(n+1) - P(n)| = |P'(\zeta)| |n+1-n|\ge 3$$ so there is an integer $k$ between $P(n)$ and $P(n+1)$.
By the intermediate value theorem you can find some real number $x_0 \in (n, n+1)$ such that $P(x_0)=k$, and you are done.
A: Short answer:
With $P(n)$ integer,
$$P(n+1)-P(n)$$ is a non-constant polynomial and will exceed $1$ for some $n$, so that in a monotonic section $P^{-1}(P(n)+1)$ cannot be an integer.
[For safety, one must ensure that $P(x)=P(n)+1$ has a unique solution. This is always achievable by taking $n$ such that $P(n)$ is larger than the largest maximum.]
A: It is true for $p$ iff it is true for $-p$ . So WLOG   assume the leading co-efficient of $p$ is positive.
Now $p$ is continuous and $p(x)\to \infty$ as $x\to \infty,$ so $\{p(x):x>r\}\supset (p(r),\infty)$ for any $r.$
And $ p'(x)\to \infty$ as $x\to \infty.$ So take $r$ such that $x>r\implies p'(x)>1.$
Take $n\in \mathbb N\cap (p(r),\infty).$ There exists $x_1>r$ with $p(x_1)=n$ and $x_2>x_1$ with $p(x_2)=n+1.$ Since $p'$ is continuous we have $$1=p(x_2)-p(x_1)=\int_{x_1}^{x_2}p'(x)\;dx>\int_{x_1}^{x_2}1\cdot dx=(x_2-x_1)>0$$ so $1>x_2-x_1>0.$ So $x_1$ and $x_2$ cannot both be integers.
A: Let's assume first that all coefficients are integers.
Let $n$ be the degree of $P$, $p > n$ a prime that does not divide the leading coefficient of $P$ and let $\overline{P}$ be the reduction of $P$ mod $p$. Now show two things:


*

*For every integer $a$ with $0 \leq a \leq p-1$, there exists an integer $b$ with $a \equiv b \mod p$ such that there exists an $x_b \in \mathbb{R}$ with $P(x_b) = b$ (hint: make $|b|$ big enough to force roots of a certain polynomial).

*If all these $x_b$ are integers, then $\overline{P}$ is the zero polynomial and hence by choice of $p$ also $P$ itself is.
Once you have done that, the general case with real coefficients is not that much harder (as long as you already know algebraic number theory): Let $\alpha_1,\alpha_2,\ldots, \alpha_m$ be the coefficients of $P$. Replace the ring of integers $\mathbb{Z}$ with the integral closure in $\mathbb{Q}(\alpha_1,\alpha_2,\ldots, \alpha_m)$ and repeat the same argument. :)
