Problem 1861

Solution:

Let us consider the polynomial $Q(x)=P(x)-1$. Condition $(*)$ ensures that $Q\left(a_{1}\right)=Q\left(a_{2}\right)=Q\left(a_{3}\right)=Q\left(a_{4}\right)=0$ and, by the Factor Theorem, $Q(x)$ is divisible by each of the factors $(x-a_{i})$ ($1 \leq i \leq 4$). Since the $a_{i}$ are distinct, $Q(x)$ is divisible by their product, that is
$$Q(x)=(x-a_{1})(x-a_{2})(x-a_{3})(x-a_{4}) R(x)$$
for some polynomial $R(x)$ with integer coefficients. If, by contradiction, we had $P(n)=12$, we would have $Q(n)=11$, that is
$$11=(n-a_{1})(n-a_{2})(n-a_{3})(n-a_{4}) R(n)$$
and therefore $11$, which is a prime number, could be written as a product of at least 4 distinct integers. But this is impossible, because the only integers that are divisors of $11$ are $\pm 1, \pm 11$, and of these at most three can be inserted in a product whose result is $11$ ($+11$ and $-11$ cannot be inserted at the same time).

The same reasoning also shows that there do not exist a polynomial $P(x)$ satisfying condition $(*)$ and an integer $n$ such that $P(n)=1998$, because one checks that $1997=1998-1$ is a prime number.