, 2003

Solution:

The answer is (B). Indeed, let $a$ and $b$ be the two integer roots of the polynomial. We then have $(x-a)(x-b)=x^{2}-(a+b)x+ab=x^{2}+p x+q$, from which:
$$a+b=-p, \quad ab=q$$
Since $q$ is a prime, from the second relation we deduce that $a, b= \pm 1$ or $a, b= \pm q$. Since $a+b$ must be negative, we have $a=-1$ and $b=-q$ (or vice versa). Hence we obtain $-1-q=-p \Rightarrow 1+q=p$, that is, $p$ is the successor of $q$. Since the only two consecutive primes are 2 and 3, we get $p=3$ and $q=2$, and therefore there is exactly one solution.

Let $a, b$ be the two possible integer roots. Then we have $x^{2}+p x+q=(x-a)(x-b)$, from which $a+b=-p$ and $ab=q$. Since $q$ is prime, from the second condition we obtain $a= \pm 1, \pm q$ and correspondingly $b= \pm q, \pm 1$. Possibly swapping $a$ and $b$, we may assume that $a= \pm 1, b= \pm q$. The first condition requires that $b$ be negative, so $a=-1, b=-q$ and $p-q=1$. The only primes differing by 1 are 2 and 3. Therefore we only have the pair (3,2), corresponding to $a=-1, b=-3$.