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Why is p squared minus one divisible by 24 for primes above 3?

Prove divisibility by coprime factors

To prove an integer is divisible by 24, it is enough to prove divisibility by 8 and by 3 separately, since 8 and 3 are coprime.

Problem

Prove that $p^{2} - 1$ is divisible by 24 whenever $p$ is a prime number greater than 3.

Answer

24

divides

p^{2} - 1

for every prime

p > 3

The restriction matters: p=2 and p=3 do not satisfy the claim.

Step-by-step solution

1. Factor the expression

Write $p^{2} - 1 = (p - 1) (p + 1)$ . Since $p > 3$ is prime, it is odd, so $p = 2 k + 1$ for an integer $k$ .

2. Show divisibility by eight

Then $(p - 1) (p + 1) = 4 k (k + 1)$ . One of two consecutive integers $k$ and $k + 1$ is even, so this product is a multiple of 8.

3. Show divisibility by three

The prime $p > 3$ is not a multiple of 3. It is therefore one less or one more than a multiple of 3, making either $p - 1$ or $p + 1$ divisible by 3. Thus $p^{2} - 1$ is divisible by both 3 and 8, hence by 24.

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