Let p be prime number such that `3 < p < 50`, then `p^2 - 1` is :

Options

A. always divisible by 8
B. always divisible by 24
C. always divisible by 12
D. all of a,b,c

Answer

To solve the problem, we need to determine the expression \( p^2 - 1 \) for prime numbers \( p \) in the range \( 3 < p < 50 \) and check its divisibility by 8, 12, and 24.

### Step-by-Step Solution:

1. **Identify the Prime Numbers**:
The prime numbers between 3 and 50 are:
\[ 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 \]

2. **Use the Formula \( p^2 - 1 \)**:
The expression \( p^2 - 1 \) can be factored as:
\[ p^2 - 1 = (p - 1)(p + 1) \]
This means we need to analyze the product of two consecutive even numbers, since \( p \) is an odd prime (all primes greater than 2 are odd).

3. **Check Divisibility by 8**:
Since \( p - 1 \) and \( p + 1 \) are two consecutive even numbers, at least one of them is divisible by 4, and the other is divisible by 2. Thus, their product \( (p - 1)(p + 1) \) is divisible by:
\[ 4 \times 2 = 8 \]

4. **Check Divisibility by 12**:
Among the two consecutive even numbers, one of them is divisible by 4, and since they are consecutive, one of them is also divisible by 2. Therefore, the product \( (p - 1)(p + 1) \) is also divisible by:
\[ 4 \times 3 = 12 \]

5. **Check Divisibility by 24**:
For divisibility by 24, we need to check if the product \( (p - 1)(p + 1) \) is divisible by \( 8 \times 3 \). Since we already established that the product is divisible by 8, we also need to ensure that one of the two even numbers is divisible by 3. Since every third odd number is prime, at least one of \( p - 1 \) or \( p + 1 \) will be divisible by 3 when \( p \) is a prime number greater than 3. Thus, \( (p - 1)(p + 1) \) is also divisible by 24.

6. **Conclusion**:
Since \( p^2 - 1 \) is divisible by 8, 12, and 24 for all prime numbers \( p \) in the specified range, we conclude that the correct answer is:
\[ \text{Option 4: all of A, B, C} \]