Find the integral solutions to the equation `[x][y]=x+y` . Show that all the non-integral solutions lie on exactly two lines. Determine these lines. Here [.] denotes greatest integer function.

Answer

To solve the equation \([x][y] = x + y\) where \([.]\) denotes the greatest integer function, we will break our solution into two cases: integral solutions and non-integral solutions.

### Step 1: Finding Integral Solutions

1. **Assume \(x\) and \(y\) are integers**:
- In this case, \([x] = x\) and \([y] = y\).
- The equation simplifies to:
\[ xy = x + y \]
- Rearranging gives:
\[ xy - x - y = 0 \quad \Rightarrow \quad xy - x - y + 1 = 1 \quad \Rightarrow \quad (x-1)(y-1) = 1 \]

2. **Finding pairs \((x-1)(y-1) = 1\)**:
- The integer pairs that satisfy this equation are:
- \((1, 1)\) which gives \((x, y) = (2, 2)\)
- \((-1, -1)\) which gives \((x, y) = (0, 0)\)

3. **Conclusion for integral solutions**:
- The integral solutions to the equation are:
\[ (0, 0) \quad \text{and} \quad (2, 2) \]

### Step 2: Finding Non-Integral Solutions

1. **Assume \(x\) and \(y\) are not integers**:
- Let \(x = k_1 + f_1\) and \(y = k_2 + f_2\) where \(k_1, k_2\) are integers and \(0 < f_1, f_2 < 1\).
- Then, \([x] = k_1\) and \([y] = k_2\).
- The equation becomes:
\[ k_1 k_2 = (k_1 + f_1) + (k_2 + f_2) \quad \Rightarrow \quad k_1 k_2 = k_1 + k_2 + f_1 + f_2 \]

2. **Rearranging the equation**:
- We can rearrange it to:
\[ k_1 k_2 - k_1 - k_2 = f_1 + f_2 \]
- Since \(f_1 + f_2\) is a sum of two fractional parts, it lies between \(0\) and \(2\).

3. **Analyzing the integer part**:
- The left-hand side \(k_1 k_2 - k_1 - k_2\) must be an integer.
- Therefore, \(f_1 + f_2\) must also be an integer, which implies:
\[ f_1 + f_2 = 1 \quad \text{(since it cannot be 0 or exceed 2)} \]

4. **Finding relationships between \(k_1\) and \(k_2\)**:
- We can express \(f_2\) in terms of \(f_1\):
\[ f_2 = 1 - f_1 \]
- Substituting back into the rearranged equation gives:
\[ k_1 k_2 = k_1 + k_2 + 1 \]
- Rearranging gives:
\[ (k_1 - 1)(k_2 - 1) = 2 \]

5. **Finding pairs \((k_1 - 1)(k_2 - 1) = 2\)**:
- The integer pairs that satisfy this equation are:
- \((1, 2)\) which gives \((k_1, k_2) = (2, 3)\)
- \((2, 1)\) which gives \((k_1, k_2) = (3, 2)\)
- \((-1, -2)\) which gives \((k_1, k_2) = (0, -1)\)
- \((-2, -1)\) which gives \((k_1, k_2) = (-1, 0)\)

6. **Finding the lines**:
- From the pairs, we can derive the lines:
- For \(k_1 = 2\) and \(k_2 = 3\), we have \(x + y = 6\).
- For \(k_1 = 3\) and \(k_2 = 2\), we have \(x + y = 6\).
- For \(k_1 = 0\) and \(k_2 = -1\), we have \(x + y = 0\).
- For \(k_1 = -1\) and \(k_2 = 0\), we have \(x + y = 0\).

### Conclusion
- The non-integral solutions lie on the lines:
\[ x + y = 6 \quad \text{and} \quad x + y = 0 \]