If $$ p $$ is a prime number greater than 3, then $$ p^2 - 1 $$ is always divisible by which number?

Question

Accepted Answer

To solve this, consider $$ p^2 - 1 $$, which can be factored as:  
$$
p^2 - 1 = (p - 1)(p + 1).
$$

Both $$ p - 1 $$ and $$ p + 1 $$ are consecutive integers surrounding the prime number $$ p $$. Since $$ p > 3 $$, it is odd, meaning both $$ p - 1 $$ and $$ p + 1 $$ are even. Therefore, their product is divisible by **4**.

Additionally, among $$ p - 1 $$, $$ p $$, and $$ p + 1 $$, one of these numbers is divisible by **3** because they are three consecutive integers. Hence, the product $$ (p - 1)(p + 1) $$ is divisible by both **4** and **3**, making it divisible by **12**.

Finally, since both $$ p - 1 $$ and $$ p + 1 $$ are even, one of them is divisible by **2**, making the product divisible by an additional factor of **2**. This results in divisibility by **24**.

Verification with examples:  
- For $$ p = 5 $$:  
$$
p^2 - 1 = 25 - 1 = 24,
$$  
which is divisible by **24**.

- For $$ p = 7 $$:  
$$
p^2 - 1 = 49 - 1 = 48,
$$  
which is divisible by **24**.

- For $$ p = 11 $$:  
$$
p^2 - 1 = 121 - 1 = 120,
$$  
which is divisible by **24**.

Thus, for any prime number greater than **3**, $$ p^2 - 1 $$ is always divisible by **24**.

Answer

Answer

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