For integral solution (x - 1) (y - 1) = 1 , Solutions are (0, 0) or (2, 2)For non integral solution, Let x = [x] + f1 and Y = [y] + f2 ∴ [x] [y] = [x] + f1 + [y] + f2 ([x] - 1) ([y] - 1) = f1+f2 + 1 Now 0 ≤ f1+f2 < 2 f1+f2 = 1 ⇒ ([x] - 1) ([y] - 1) = 2 Which is possible for [x] = 3 and [y] = 2or [x] = 2 and [y] = 3 or [x] = - 1 and [y] = 0or [x] = 0 and [y] = - 1 ∴ The x + y = [x] [y] becomes x + y = 6 or x + y = 0 ∴ Non integral solution lies on x + y = 6 or x + y = 0. No. of point of intersections is clearly zero