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
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.
Let's assume first that all coefficients are integers.

Let $n$ be the degree of $P$, $p &gt; n$ 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. :)