Calculation:Let y3 + ay2 + by + c = f(y)? If f(y) is divisible by y2 + 1, then by putting y2 + 1 = 0 ? or y2 = “1, f(y) should become 0.? y2.y + ay2 + by + c = 0(for y2 = -1)? or, “y “ a + by + c = 0? or, y(b “ 1) + (c “ a) = 0 ¦(i)? For (i) to be true, b = 1 and c = a.? Now, c (= a) can take any of the six values 1, 2, 3, 4, 5 and 6.? Number of such polynomials is 6.Alternate MethodAs y2 + 1 divides y3 + ay2 + by + c,? y3 + ay2 + by + c = (y2 + 1)(y + c)? Or, y3 + ay2 + by + c = y3 + cy2 + y + c.? Hence, b = 1 and a = c? Now, c (= a) can take any of the six values 1, 2, 3, 4, 5 and 6.? Number of such polynomials is 6