The sum of three consecutive integers is equal to their product. How many such possibilities are there?

Re-checking: let the integers be n-1, n, n+1. Sum = 3n and product = n(n-1)(n+1)=n(n^2-1). Equating: 3n = n(n^2-1) ⇒ n(n^2-4)=0 ⇒ n = 0, 2, -2.

• n = 0 gives (-1, 0, 1); sum = 0, product = 0. Correct.
• n = 2 gives (1, 2, 3); sum = 6, product = 6. Correct.
• n = -2 gives (-3, -2, -1); sum = -6, product = -6. Correct.

Hence there are three possibilities. My earlier derived answer (b) was an error; the correct answer is (c).