The sum of all values of integers n for which `(n^2-9)/(n-1)` is also an integer is

Explanation

To solve the problem of finding the sum of all integer values of \( n \) for which the expression \( \frac{n^2 - 9}{n - 1} \) is also an integer, we can follow these steps:

### Step 1: Simplify the Expression
We start with the expression:
\[ \frac{n^2 - 9}{n - 1} \]
We can factor \( n^2 - 9 \) as:
\[ n^2 - 9 = (n - 3)(n + 3) \]
Thus, we rewrite the expression:
\[ \frac{(n - 3)(n + 3)}{n - 1} \]

### Step 2: Rewrite the Expression
Next, we can express \( n^2 - 9 \) in a different form:
\[ n^2 - 9 = n^2 - 1 - 8 = (n^2 - 1) - 8 \]
This allows us to break it down:
\[ \frac{n^2 - 1 - 8}{n - 1} = \frac{n^2 - 1}{n - 1} - \frac{8}{n - 1} \]
Now, we know that:
\[ n^2 - 1 = (n - 1)(n + 1) \]
So we can simplify:
\[ \frac{(n - 1)(n + 1)}{n - 1} - \frac{8}{n - 1} = n + 1 - \frac{8}{n - 1} \]

### Step 3: Determine Conditions for Integer Values
For \( n + 1 - \frac{8}{n - 1} \) to be an integer, \( \frac{8}{n - 1} \) must also be an integer. This means that \( n - 1 \) must be a divisor of 8.

### Step 4: Find Divisors of 8
The divisors of 8 are:
\[ \pm 1, \pm 2, \pm 4, \pm 8 \]
Thus, we can set \( n - 1 \) equal to each of these divisors to find possible values of \( n \).

### Step 5: Calculate Possible Values of \( n \)
1. If \( n - 1 = 1 \) then \( n = 2 \)
2. If \( n - 1 = -1 \) then \( n = 0 \)
3. If \( n - 1 = 2 \) then \( n = 3 \)
4. If \( n - 1 = -2 \) then \( n = -1 \)
5. If \( n - 1 = 4 \) then \( n = 5 \)
6. If \( n - 1 = -4 \) then \( n = -3 \)
7. If \( n - 1 = 8 \) then \( n = 9 \)
8. If \( n - 1 = -8 \) then \( n = -7 \)

The possible integer values of \( n \) are:
\[ n = 2, 0, 3, -1, 5, -3, 9, -7 \]

### Step 6: Sum the Values
Now, we sum all these values:
\[ 2 + 0 + 3 - 1 + 5 - 3 + 9 - 7 = 8 \]

### Final Answer
The sum of all integer values of \( n \) for which \( \frac{n^2 - 9}{n - 1} \) is an integer is:
\[ \boxed{8} \]