This basically says that for some $N$, for all $x > N$, $x$ can be written as $P(n)$ for some integer $n$ For any polynomial of degree greater than $1$, its well known that we can get the difference between $P(x+1)$ and $P(x)$ to be arbitrarily large for large enough $x$ and so the degree of $P(x)$ must be $\le 1$. Constant polynomials dont work, and obviously if $P(x) = ax+b$ for some real numbers $a,b$. Obviously $a \le 1$ since otherwise it will miss infinitely many numbers. Also, $a,b$ must be rational. So, $P(x) = \frac{px+q}{r}$ for some $p \le r$. But this means that only finitely many numbers $n$ exist such that $rn \equiv q \pmod p$, this implies that $p | r$ and so if $r = pk$, $P(x) = \frac{x}{k} + \frac{q}{pk}$, but if $P(x) = n$, it means we need a value of $x = \frac{npk - q}{p}$, but since $x$ must be an integer, $p | q$ and so let $q = cp$ which means that $P(x) = \frac{x+c}{k}$ Therefore, the only solution is $P(x) = \frac{x+c}{k}$, which obviously works @below - Thanks for pointing that out... I didnt see that, i thought $P(x)$ was an integer polynomial

L567 wrote: Therefore, the only solution is $P(x) = x+c$, which obviously works Note that the polynomials have real coefficients. The solution set I get is $\dfrac{x+c}{d}$, where $c, d \in \mathbb{Z}$ and $d \neq 0$.