Prove that p^2 - 1 is divisible by 24 for any prime number p >= 5

This is one of the basic problems in understanding the properties of prime numbers and the divisibility rule. There could be several ways to prove this.

Proof idea 1

As we know, P^2 - 1 = (p + 1)(p - 1). On the other side, the multiplication of three consecutive numbers is divisible by 3. Since p is a prime number, (p - 1)(p + 1) is divisible by 3.

Now assume, p = 2n + 1, then (p + 1)(p - 1) = 2n(2n + 2) = 4n(n + 1). Multiplication of two consecutive numbers is divisible by 2. So n(n + 1) is divisible by 2 => 4n(n + 1) is divisible by 8. Overall, (p + 1)(p - 1) is divisible by 3 and 8 and both are relatively prime => (p + 1)(p - 1) is divisible by 3*8 i.e. 24.

Critical ideas to explore:

• Let A, B, and C be three integers, such that A divides C, B divides C. If A and B are relatively prime, then AB divides C.
• P^2 - 1 will be also divisible by 24 if p is an odd number not multiple of 3.

Proof idea 2

We can write every prime number p greater than 5 in the form of 6n + 1 or 6n - 1. Why? Here is an idea: Any integer can be expressed as 6n, 6n + 1, 6n - 1, 6n + 2, 6n - 2, 6n + 3. Out of these six possibilities: 6n, 6n + 2, 6n - 2 and 6n + 3 can not be prime numbers. So there are two possibilities to express prime numbers: 6n + 1 or 6n - 1.

• If p = 6n + 1, p^2 - 1 = (p + 1)(p - 1) = (6n + 2)(6n) = 12n(3n + 1).
• If p = 6n - 1, p^2 - 1 = (p + 1)(p - 1) = (6n)(6n - 2) = 12n(3n - 1).

If n is even then 12n will be divisible by 24. So, 12n(3n + 1) and 12n(3n - 1) will be divisible by 24. If n is odd, then (3n + 1) or (3n - 1) will be even and 12(3n + 1) or 12(3n - 1) will be divisible by 24. So overall, 12n(3n + 1) and 12n(3n - 1) will be divisible by 24.

Proof idea 3

As we know, the sum of squares of the first n positive numbers = n*(n + 1)(2n + 1)]/6. Now, if p (> 3) is a prime then (p - 1)/2 will be a positive integer. Now, if we put n = (p - 1)/2 in the sum formula, we will get the value of sum = [p*(p^2 - 1)]/24. This value must be an integer.

Here p and 24 are relatively prime i.e. gcd (p, 24) = 1. So p^2 - 1 will be divisible by 24.

Critical ideas to think!

• Can we think of proving it using some other method?
• Here is a more difficult question: Prove that if p is a prime greater than 7, then p^4 - 1 is divisible by 240.
• If an integer is simultaneously a square and a cube (ex: 64 = 82 = 43 ), verify that the integer must be of form 7n or 7n + 1.
• Prove by induction: For n ≥ 1, 8^n − 3^n is divisible by 5.
• Prove by induction: a number, given its decimal representation, is divisible by 3 if the sum of its digits is divisible by three.

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