Let $ m, n$ be the roots of the original equation (not necessarily distinct) WLOG, let $ m\geq n$ Then $ m=\frac{a+\sqrt{a^2-4a}}{2}, n=\frac{a-\sqrt{a^2-4a}}{2}$. Since $ m, n$ are integers, $ m-n = \sqrt{a^2-4a}$ is also a positive integer, name it as $ k$. So $ a=2m-k$ is also an integer. Now, $ a^2-4a = k^2$ $ (a-2)^2 - k^2 = 4$ $ (a-2+k)(a-2-k) = 4$, as $ a,k$ are integers, $ (a-2+k)$, $ (a-2-k)$ have the same parity. So (a-2+k) = 2, (a-2-k) = 2 or (a-2+k) = -2, (a-2-k) = -2. Solve the two simultaneous equations we can get $ (a,k) = (4,0)$ or $ (0,0)$. So $ a = 0, 4$