We take a in the form $2k+\frac{3}{2}$ where k is an integer. We will find that it works

I claim that numbers of the form $\frac{k}{2}$, where $k \equiv 3 \pmod{4}$ work. Note that ${a}=\frac{1}{2}$ because $k$ is odd. Therefore, the expression simplifies to $\frac{k(k-3)}{2}$ which is obviously an integer, by our hypothesis. $\blacksquare$