## Prime Uniqueness

#### Problem

Prove that seven is the only prime number that is one less than a perfect cube.

#### Solution

Let `p` be a prime number one less than a perfect cube, `p` = `n`^{3} 1

By factoring the right hand side,`p` = (`n` 1)(`n`^{2} + `n`+ 1)

By definition `p` cannot have any factors, so `n` 1 = 1 `n` = 2.

Hence `p` = `2`^{3} 1 = 7.

Investigate this property for other perfect powers.

Problem ID: 48 (May 2001) Difficulty: 2 Star