We have $$-1<a^{2}-D b^{2}<1 \Rightarrow \frac{a^{2}-1}{b^{2}}<D<\frac{a^{2}+1}{b^{2}}$$ We have $\frac{a^{2}-1}{b^{2}}$ is maximal at $-15=\frac{.25^{2}-1}{.25^{2}}$ and $\frac{a^{2}+1}{b^{2}}$ is minimal at $\frac{0^{2}+1}{.25^{2}}=16$. However, since we cannot have $a, b= \pm .25$, checking border cases of -15 and 16 shows that both of these values are possible for $D$. Hence, $-15 \leq D \leq 16$, so there are 32 possible values of $D$.