A2. Let $p$ and $q$ be different prime numbers. For how many different values of $a$ are both solutions of the quadratic equation $x^{2}+a x+p q=0$ integers?
(A) 4
(B) 3
(C) 2
(D) 1
(E) 0
A2. Since the coefficient of $x^{2}$ is 1, by Vieta's formulas, the product $x_{1} x_{2}$ of both solutions of the equation is equal to $p q$. Since both solutions must be integers and $p$ and $q$ are prime numbers, the set of solutions $\left\{x_{1}, x_{2}\right\}$ has only the possibilities $\{1, p q\},\{-1,-p q\},\{p, q\}$ or $\{-p,-q\}$. By Vieta's formulas, the coefficient $a$ is equal to $-\left(x_{1}+x_{2}\right)$, so it can have 4 different values, which are $1+p q,-1-p q, p+q$ and $-p-q$. All 4 values are distinct from each other.