`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 the values of \( n \) for which \( n^2 - 1 \) is divisible by \( 8 \), we can analyze the expression step by step.

### Step 1: Understand 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 \( n + 1 \).

### Step 2: Analyze the product of two consecutive integers
The product of two consecutive integers is always even. In fact, one of them is guaranteed to be divisible by \( 2 \) and at least one of them will be divisible by \( 4 \) if the integers are consecutive. Therefore, the product \( (n - 1)(n + 1) \) is divisible by \( 8 \) if one of these integers is divisible by \( 4 \) and the other is even.

### Step 3: Consider the cases for \( n \)
1. **Case 1: \( n \) is odd**
- If \( n \) is odd, we can express \( n \) as \( n = 2k + 1 \) for some integer \( k \).
- Then, \( n - 1 = 2k \) (even) and \( n + 1 = 2k + 2 \) (even).
- The product \( (n - 1)(n + 1) = 2k(2k + 2) = 4k(k + 1) \).
- Since \( k(k + 1) \) is the product of two consecutive integers, it is always even. Thus, \( 4k(k + 1) \) is divisible by \( 8 \).

2. **Case 2: \( n \) is even**
- If \( n \) is even, we can express \( n \) as \( n = 2k \) for some integer \( k \).
- Then, \( n - 1 = 2k - 1 \) (odd) and \( n + 1 = 2k + 1 \) (odd).
- The product \( (n - 1)(n + 1) = (2k - 1)(2k + 1) \) is the product of two odd numbers, which is odd and hence not divisible by \( 8 \).

### Conclusion
From the analysis, we conclude that \( n^2 - 1 \) is divisible by \( 8 \) if and only if \( n \) is an odd number.

### Final Answer
Thus, \( n \) must be an odd number.

---