Solve `x^2-4-[x]=0` (where `[]` denotes the greatest integer function).

Answer

To solve the equation \( x^2 - 4 - [x] = 0 \), where \([x]\) denotes the greatest integer function, we can follow these steps:

### Step 1: Rewrite the equation
We can rewrite the equation as:
\[ x^2 - 4 = [x] \]
This means we need to find the values of \(x\) such that the left-hand side equals the greatest integer of \(x\).

### Step 2: Analyze the left-hand side
The left-hand side \(x^2 - 4\) is a quadratic function. It opens upwards (since the coefficient of \(x^2\) is positive) and has roots at:
\[ x^2 - 4 = 0 \implies x = 2 \quad \text{and} \quad x = -2 \]
The vertex of the parabola occurs at:
\[ x = 0 \quad \text{(since the vertex of } ax^2 + bx + c \text{ is at } x = -\frac{b}{2a} \text{ and here } b = 0) \]
At \(x = 0\):
\[ x^2 - 4 = 0^2 - 4 = -4 \]

### Step 3: Determine the range of \(x^2 - 4\)
The function \(x^2 - 4\) takes values from \(-4\) (at \(x=0\)) to \(+\infty\) as \(x\) moves away from zero. Thus, the range of \(x^2 - 4\) is \([-4, \infty)\).

### Step 4: Analyze the greatest integer function \([x]\)
The greatest integer function \([x]\) takes integer values. For any real number \(x\), \([x]\) is the largest integer less than or equal to \(x\). Therefore, we need to find integer values that \(x^2 - 4\) can equal.

### Step 5: Set up intervals
We will consider intervals based on the integer values of \([x]\):
1. For \([x] = n\), where \(n\) is an integer, we have:
\[ x^2 - 4 = n \implies x^2 = n + 4 \implies x = \pm\sqrt{n + 4} \]
We need to ensure that \(n \leq x < n + 1\) for the corresponding \(x\).

### Step 6: Solve for various integer values of \(n\)
1. **For \(n = -4\)**:
\[ x^2 = 0 \implies x = 0 \quad \text{(valid since } -4 \leq 0 < -3\text{)} \]

2. **For \(n = -3\)**:
\[ x^2 = 1 \implies x = \pm 1 \quad \text{(valid since } -3 \leq 1 < -2\text{)} \]

3. **For \(n = -2\)**:
\[ x^2 = 2 \implies x = \pm\sqrt{2} \quad \text{(valid since } -2 \leq \sqrt{2} < -1\text{)} \]

4. **For \(n = -1\)**:
\[ x^2 = 3 \implies x = \pm\sqrt{3} \quad \text{(valid since } -1 \leq \sqrt{3} < 0\text{)} \]

5. **For \(n = 0\)**:
\[ x^2 = 4 \implies x = \pm 2 \quad \text{(valid since } 0 \leq 2 < 1\text{)} \]

6. **For \(n = 1\)**:
\[ x^2 = 5 \implies x = \pm\sqrt{5} \quad \text{(valid since } 1 \leq \sqrt{5} < 2\text{)} \]

7. **For \(n = 2\)**:
\[ x^2 = 6 \implies x = \pm\sqrt{6} \quad \text{(valid since } 2 \leq \sqrt{6} < 3\text{)} \]

8. **For \(n = 3\)**:
\[ x^2 = 7 \implies x = \pm\sqrt{7} \quad \text{(valid since } 3 \leq \sqrt{7} < 4\text{)} \]

### Step 7: Compile the solutions
From the valid intervals, we find:
- \(x \in (-2, -1)\) from \(n = -2\)
- \(x \in (2, 3)\) from \(n = 2\)

Thus, the final solution is:
\[ x \in (-2, -1) \cup (2, 3) \]