mathschallenge.net

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.