A special polynomial!

Let $P(n)=a_k n^k + a_{k-1} n^{k-1} + \cdots +a_1 n+a_0$ . Consider $P$ evaluated at two distinct integer values, $n$ and $r$ : $P(n)-P(r)=a_k \left(n^k-r^k \right)+a_{k-1} \left(n^{k-1}-r^{k-1} \right)+\cdots a_1 \left(n-r \right)$

Note that every term on the right-hand side is divisible by $n-r$ ; so $\frac{P(n)-P(r)}{n-r}$ is an integer.

Now assume that $r$ is an integer root of $P$ , ie $P(r)=0$ , and let $n=1$ . Then $1-r|P(1)$ Similarly, $2018-r|P(2018)$

But both $P(1)$ and $P(2018)$ are three-digit integers, and at least one of $1-r$ and $2018-r$ must be outside the range $[-999,999]$ ; contradiction.

So $P(n)$ has no integer roots.