Question

A positive integer is said to be a prime number if it is not divisible by any positive integer other than itself and 1. Let p be a prime number greater than 5. Then (p2 - 1) is

A) always divisible by 6, and may or may not be divisible by 12
B) always divisible by 24
C) never divisible by 6
D) always divisible by 12, and may or may not be divisible by 24

Answer 1

Answer:

The correct option is B.

Explanation:

If p is a prime number greater than 3, then p2-1 is always divisible by 24.

p2 - 1 = (p - 1) x (p + 1)

As p is a prime number, it must be odd. So, p - 1 and p + 1 must be even. Now, these 2 even numbers are consecutive. One of them must be a multiple of 4.

p - 1, p and p + 1 form three consecutive numbers. In any three consecutive numbers, one will be a multiple of 3. As p is not a multiple of 3 (p is prime), hence either p - 1 or p + 1 is a multiple. Therefore p2 - 1 has factors: 2, 4, & 3.  

Hence, p2 - 1 = 24n

When p = 7, p2 - 1 = 48

When p = 11, p2 - 1 = 120

Answer 2

Given:

Prime number, $\[p > 5\]$

To Find: $\[{p^2} - 1\]$

Solution:

For every prime number which is greater than 3, the value of $\[{p^2} - 1\]$ is always divisible by 24.

The expression $\[{p^2} - 1\]$ can be represented as:

$\[{p^2} - 1 = (p - 1)(p + 1)\]$

As we know, a prime number greater than 2 is always an odd number.

Therefore, the value of $\[p - 1\]$ and $\[p + 1\]$ must be an even number.

These two even numbers are consecutive even numbers. This shows that one of them should be a multiple of 4.

Now, the three numbers, i.e., $\[(p - 1),\,p,\,(p + 1)\]$ are three consecutive numbers, this shows that any one of these numbers should be a multiple of 3.

The prime number $p$ is never a multiple of 3, therefore, either $\[p - 1\]$ or $\[p + 1\]$

Therefore, the factors of $\[{p^2} - 1\]$ are 2, 3 and 4.

Thus, it is always divisible by 24.

Hence, $\[{p^2} - 1\]$ is always divisible by 24 and the correct option is B) always divisible by 24.

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