Balti Tee 1998 · Ülesanne 4

Solution:

Let $m$ be an arbitrary integer and define $n \in \{1,2, \ldots, 1998\}$ to be such that $m \equiv n \pmod{1998}$. Then $P(m) \equiv P(n) \pmod{1998}$. Since $P(n)$ as a three-digit number cannot be divisible by $1998$, then $P(m)$ cannot be equal to $0$. Hence $P$ has no integer roots.