## Consecutive Cube

#### Problem

Three consecutive integers are multiplied together and the middle number is added.

E.g. 3 4 5 = 60 and 60 + 4 = 64 = 4^{3}

Will this always produce a cube number?

#### Solution

Taking three consecutive integers, `n`, `n` + 1 and `n` + 2 we get.

n(n + 1)(n + 2) + (n + 1) | = (n + 1)(n(n + 2) + 1) |

= (n + 1)(n^{2} + 2n + 1) | |

= (n + 1)(n + 1)^{2} | |

= (n + 1)^{3} |

Problem ID: 83 (Oct 2002) Difficulty: 2 Star