What would be the values of n for which `n^(2)-1` is divisible by 8.

Answer

To determine the values of \( n \) for which \( n^2 - 1 \) is divisible by 8, we can follow these steps:

### Step 1: Understand the expression
We need to find when \( n^2 - 1 \) is divisible by 8. We can rewrite \( n^2 - 1 \) as:
\[ n^2 - 1 = (n - 1)(n + 1) \]
This means we need to check when the product of \( (n - 1) \) and \( (n + 1) \) is divisible by 8.

### Step 2: Analyze the parity of \( n \)
Since \( n \) can be either even or odd, we will consider both cases.

#### Case 1: \( n \) is even
Let \( n = 2k \) (where \( k \) is an integer). Then:
\[ n - 1 = 2k - 1 \quad \text{and} \quad n + 1 = 2k + 1 \]
Both \( n - 1 \) and \( n + 1 \) are odd numbers. The product of two odd numbers is odd, and thus cannot be divisible by 8.

#### Case 2: \( n \) is odd
Let \( n = 2k + 1 \) (where \( k \) is an integer). Then:
\[ n - 1 = 2k \quad \text{and} \quad n + 1 = 2k + 2 \]
Here, \( n - 1 \) is even and \( n + 1 \) is also even. The product \( (n - 1)(n + 1) \) will be:
\[ (2k)(2k + 2) = 2k \cdot 2(k + 1) = 4k(k + 1) \]
For \( 4k(k + 1) \) to be divisible by 8, \( k(k + 1) \) must be even. Since \( k \) and \( k + 1 \) are consecutive integers, one of them is always even, thus \( k(k + 1) \) is always even.

### Step 3: Conclusion
Since \( n \) must be odd for \( n^2 - 1 \) to be divisible by 8, we can conclude that:
- The values of \( n \) for which \( n^2 - 1 \) is divisible by 8 are all odd integers.

### Summary
The values of \( n \) for which \( n^2 - 1 \) is divisible by 8 are all odd integers.