, 2007

Solution:

The answer is $19$. Subtracting the two equations term by term we get $x^{2} - y^{2} = y - x$, that is $(x - y)(x + y + 1) = 0$.

Suppose that the first factor vanishes, that is $y = x$: substituting, $x$ must satisfy the equation $x^{2} - x - a = 0$. Hence $a$ must be such that
$$x = \frac{1 \pm \sqrt{1 + 4a}}{2} \text{ is an integer. }$$
This happens if and only if $a$ is such that $1 + 4a$ is the square of an odd number. In other words, $a$ must be such that there exists an $n \in \mathbb{N}$ such that $1 + 4a = (2n + 1)^{2}$, that is such that $a = n(n + 1)$. Since $0 < a < 100$, the acceptable values of $n$ are those from $1$ to $9$ inclusive.

Now suppose that the second factor vanishes. Proceeding analogously to the previous case, $y = -x - 1$; substituting, $x$ must satisfy the equation $x^{2} + x + (1 - a) = 0$. $a$ must be such that
$$x = \frac{-1 \pm \sqrt{1 + 4(a - 1)}}{2} \text{ is an integer. }$$
Once again $1 + 4(a - 1)$ must be the square of an odd integer, that is $1 + 4(a - 1) = (2n + 1)^{2}$, that is $a = n(n + 1) + 1$. The acceptable values of $n$ are those from $0$ to $9$ inclusive and give all values of $a$ distinct from those of the previous case.

Summing up we have $9$ values of $a$ from the first case and $10$ from the second for a total of $19$ integer values of $a$ for which the system has integer solutions.