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.
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:
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 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.
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.
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There are 3 doors behind which are two goats and a car. You pick door 1 hoping for the car but don’t open it right away. Monty Hall, the game show host who knows what's behind the doors, opens door 3, which has a goat. Here's the game: do you want to pick door No. 2? Is it to your advantage to switch your choice?
Given two non-negative integers, m and n, we have to find their greatest common divisor or HCF. It is the largest number, a divisor of both m and n. The Euclidean algorithm is one of the oldest and most widely known methods for computing the GCD of two integers.