Find the integral solution for `n_(1)n_(2)=2n_(1)-n_(2), " where " n_(1),n_(2) in "integer"`.

Answer

To find the integral solutions for the equation \( n_1 n_2 = 2 n_1 - n_2 \), where \( n_1 \) and \( n_2 \) are integers, we can follow these steps:

### Step 1: Rearranging the Equation
Start with the given equation:
\[ n_1 n_2 = 2 n_1 - n_2 \]
We can rearrange this by moving \( -n_2 \) to the left side:
\[ n_1 n_2 + n_2 = 2 n_1 \]

### Step 2: Factoring Out \( n_2 \)
Now, we can factor out \( n_2 \) from the left side:
\[ n_2 (n_1 + 1) = 2 n_1 \]

### Step 3: Solving for \( n_2 \)
Next, we can solve for \( n_2 \):
\[ n_2 = \frac{2 n_1}{n_1 + 1} \]

### Step 4: Finding Integer Solutions
We need to find values of \( n_1 \) such that \( n_2 \) is also an integer. We will check various integer values for \( n_1 \):

1. **For \( n_1 = 0 \)**:
\[ n_2 = \frac{2 \cdot 0}{0 + 1} = 0 \]
Solution: \( (0, 0) \)

2. **For \( n_1 = 1 \)**:
\[ n_2 = \frac{2 \cdot 1}{1 + 1} = \frac{2}{2} = 1 \]
Solution: \( (1, 1) \)

3. **For \( n_1 = -2 \)**:
\[ n_2 = \frac{2 \cdot (-2)}{-2 + 1} = \frac{-4}{-1} = 4 \]
Solution: \( (-2, 4) \)

4. **For \( n_1 = -3 \)**:
\[ n_2 = \frac{2 \cdot (-3)}{-3 + 1} = \frac{-6}{-2} = 3 \]
Solution: \( (-3, 3) \)

5. **For \( n_1 = 2 \)**:
\[ n_2 = \frac{2 \cdot 2}{2 + 1} = \frac{4}{3} \]
Not an integer.

6. **For \( n_1 = -1 \)**:
\[ n_2 = \frac{2 \cdot (-1)}{-1 + 1} = \frac{-2}{0} \]
Undefined.

### Step 5: Listing All Integral Solutions
After checking various integer values for \( n_1 \), we find the integral solutions:
- \( (0, 0) \)
- \( (1, 1) \)
- \( (-2, 4) \)
- \( (-3, 3) \)

### Final Answer
The integral solutions for the equation \( n_1 n_2 = 2 n_1 - n_2 \) are:
\[ (0, 0), (1, 1), (-2, 4), (-3, 3) \]