`n^(2)-1` is divisible by `8`, if `n` is

Options

A. an integer
B. a natural number
C. an odd number
D. an even number

Explanation

To determine when \( n^2 - 1 \) is divisible by \( 8 \), we can analyze the expression based on whether \( n \) is even or odd.

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression \( n^2 - 1 \). We can rewrite it as:
\[ n^2 - 1 = (n - 1)(n + 1) \]
This shows that \( n^2 - 1 \) is the product of two consecutive integers \( (n - 1) \) and \( (n + 1) \).

2. **Case 1: \( n \) is Even**:
- If \( n \) is even, we can express \( n \) as \( n = 2k \) for some integer \( k \).
- Then, we have:
\[ n^2 - 1 = (2k)^2 - 1 = 4k^2 - 1 \]
- Now, we check the divisibility by \( 8 \):
- For \( k = 0 \): \( 4(0)^2 - 1 = -1 \) (not divisible by \( 8 \))
- For \( k = 1 \): \( 4(1)^2 - 1 = 3 \) (not divisible by \( 8 \))
- For \( k = 2 \): \( 4(2)^2 - 1 = 15 \) (not divisible by \( 8 \))
- Thus, if \( n \) is even, \( n^2 - 1 \) is not divisible by \( 8 \).

3. **Case 2: \( n \) is Odd**:
- If \( n \) is odd, we can express \( n \) as \( n = 2k + 1 \) for some integer \( k \).
- Then, we have:
\[ n^2 - 1 = (2k + 1)^2 - 1 = 4k^2 + 4k + 1 - 1 = 4k^2 + 4k \]
- This can be factored as:
\[ n^2 - 1 = 4k(k + 1) \]
- Since \( k \) and \( k + 1 \) are consecutive integers, one of them is even, making \( 4k(k + 1) \) divisible by \( 8 \).
- For example:
- For \( k = 0 \): \( 4(0)(1) = 0 \) (divisible by \( 8 \))
- For \( k = 1 \): \( 4(1)(2) = 8 \) (divisible by \( 8 \))
- For \( k = 2 \): \( 4(2)(3) = 24 \) (divisible by \( 8 \))

4. **Conclusion**:
Therefore, \( n^2 - 1 \) is divisible by \( 8 \) when \( n \) is odd.

### Final Answer:
\( n \) must be an odd integer for \( n^2 - 1 \) to be divisible by \( 8 \).