We are given a polynomial $f$ with integer coefficients such that for 4 distinct integers $a_1,a_2,a_3$ and $ a_4$, $f(a_1)=f(a_2)=f(a_3)=f(a_4)=1991$. Show that there exists no integer $b$ such that $f(b)=1993$.
There must be a polynomial with integer coefficients say $q(x)$ with$$f(x)=(x-a_1)(x-a_2)(x-a_3)(x-a_4)q(x)+1991$$
If $f(b)=1993$ then $$1993=(b-a_1)(b-a_2)(b-a_3)(b-a_4)q(b)+1991$$
Since $b,q(b),a_1 ,a_2,a_3,a_4$ are distinct integers this means that 2 can be written as a product of at least 4 different integers which is a contradiction.