Baltic Way 1997 · Problem 7

Solution:

Suppose $b$ is an integer such that $Q(P(b))=1$. Since $a$ and $a+1997$ are roots of $P$ we have $P(x) = (x-a)(x-a-1997) R(x)$ where $R$ is a polynomial with integer coefficients. For any integer $b$ the integers $b-a$ and $b-a-1997$ are of different parity and hence $P(b) = (b-a)(b-a-1997) R(b)$ is even. Since $Q(1998)=2000$ then the constant term in the expansion of $Q(x)$ is even (otherwise $Q(x)$ would be odd for any even integer $x$), and $Q(c)$ is even for any even integer $c$. Hence $Q(P(b))$ is also even and cannot be equal to $1$.