## Prime Square Divisibility

#### Problem

Prove that `p`^{2}1 is divisible by 24 for all primes, `p` > 3.

#### Solution

For all primes greater than 3, `p` = 6`k`1.

When `p` = 6`k`+1, `p`^{2} = 36`k`^{2} + 12`k` + 1 and so `p`^{2}1 = 36`k`^{2} + 12`k`.

Similarly, when `p` = 6`k`1, `p`^{2}1 = 36`k`^{2} 12`k`.

Therefore, `p`^{2}1 = 36`k`^{2} 12`k` = 12`k`(3`k` 1).

If `k` is odd, 3`k` 1 will be even and so we prove that `p`^{2}1 will always be divisible by 12 2 = 24.

What can you say about `p`^{3}1?

What about other powers of `p`?

Problem ID: 67 (Feb 2002) Difficulty: 2 Star