Prove that for all prime numbers p > 3, (p^{2} – 1) is a multiple of 24.

#### Solution

First, factor the expression: p^{2} – 1 = (p + 1) (p – 1)

Among any 3 consecutive numbers, one must be a multiple of 3. Since p is a prime number greater than 3, p is not a multiple of 3. Therefore, either p + 1 or p – 1 must be a multiple of 3, which means (p + 1) (p – 1) is a multiple of 3.

Since p is a prime number greater than 3, p is odd and both p + 1 and p – 1 must be even. Among any two consecutive even numbers, one must be a multiple of 4 and the other must be a multiple of 2. Therefore (p + 1) (p – 1) is a multiple of 8.

Since we have shown (p + 1) (p – 1) is a multiple of both 3 and 8, it must be a multiple 24.