Let a = 4 + f, where 0 < f < 1. We are given that (4 + f)(4 – 2f) is an integer. This implies that 2f^{2} + 4f is an integer. Since 0 < f < 1, we have 0 < 2f^{2} + 4f < 6. Therefore 2f^{2} + 4f can be take 1, 2, 3, 4 or 5. Equating 2f^{2 }+ 4f to each one of them and using f > 0, we get