14. If $a$ is a positive integer less than 100, for how many values of $a$ does the system $\left\{\begin{array}{l}x^{2}=y+a \\ y^{2}=x+a\end{array}\right.$ have integer solutions?
14. The answer is 19. By subtracting the two equations member by member, we get \(x^{2}-y^{2}=y-x\), or equivalently \((x-y)(x+y+1)=0\).

Suppose the first factor is zero, i.e., \(y=x\): substituting, \(x\) must satisfy the equation \(x^{2}-x-a=0\). Therefore, \(a\) must be such that

\[
x=\frac{1 \pm \sqrt{1+4 a}}{2} \text{ is an integer. }
\]

This happens if and only if \(a\) is such that \(1+4 a\) is the square of an odd number. In other words, \(a\) must be such that there exists an \(n \in \mathbb{N}\) such that \(1+4 a=(2 n+1)^{2}\), or equivalently such that \(a=n(n+1)\). Since \(0<a<100\), the acceptable values of \(n\) are from 1 to 9 inclusive.

Now suppose the second factor is zero. Proceeding similarly to the previous case, \(y=-x-1\); substituting, \(x\) must satisfy the equation \(x^{2}+x+(1-a)=0\). \(a\) must be such that

\[
x=\frac{-1 \pm \sqrt{1+4(a-1)}}{2} \text{ is an integer. }
\]

Again, \(1+4(a-1)\) must be the square of an odd integer, i.e., \(1+4(a-1)=(2 n+1)^{2}\), or equivalently \(a=n(n+1)+1\). The acceptable values of \(n\) are from 0 to 9 inclusive and provide all distinct values of \(a\) from those of the previous case.

In summary, there are 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.