The solution set of the equation
`(x)^(2)+[x]^(2)=(x-1)^(2)+[x+1]^(2)`,
where (x) denotes the least integer greater than or equal to x and [x] denotes the greatest integer less than or equal to x, is

Options

A. R
B. R-Z
C. R-N
D. None of these

Explanation

To solve the equation \( (x)^2 + [x]^2 = (x-1)^2 + [x+1]^2 \), where \( (x) \) denotes the least integer greater than or equal to \( x \) (the ceiling function) and \( [x] \) denotes the greatest integer less than or equal to \( x \) (the floor function), we will analyze the equation step by step.

### Step 1: Understanding the Functions
We know that:
- \( (x) = \lceil x \rceil \) (the smallest integer greater than or equal to \( x \))
- \( [x] = \lfloor x \rfloor \) (the largest integer less than or equal to \( x \))

### Step 2: Rewrite the Equation
The equation can be rewritten as:
\[ \lceil x \rceil^2 + \lfloor x \rfloor^2 = (x-1)^2 + \lfloor x+1 \rfloor^2 \]

### Step 3: Analyze Cases
Since \( x \) can be expressed in terms of its integer and fractional parts, we can consider two cases based on the integer part \( n \) of \( x \):
1. When \( x \) is an integer.
2. When \( x \) is not an integer.

#### Case 1: \( x = n \) (where \( n \) is an integer)
In this case, we have:
- \( \lceil x \rceil = n \)
- \( \lfloor x \rfloor = n \)
- \( \lfloor x + 1 \rfloor = n + 1 \)

Substituting these into the equation gives:
\[ n^2 + n^2 = (n-1)^2 + (n+1)^2 \]
This simplifies to:
\[ 2n^2 = (n^2 - 2n + 1) + (n^2 + 2n + 1) \]
\[ 2n^2 = 2n^2 + 2 \]
This leads to:
\[ 0 = 2 \]
which is a contradiction. Thus, there are no solutions in this case.

#### Case 2: \( x = n + k \) (where \( n \) is an integer and \( 0 < k < 1 \))
In this case:
- \( \lceil x \rceil = n + 1 \)
- \( \lfloor x \rfloor = n \)
- \( \lfloor x + 1 \rfloor = n + 1 \)

Substituting these into the equation gives:
\[ (n + 1)^2 + n^2 = (n + k - 1)^2 + (n + 1)^2 \]
Expanding both sides:
\[ (n^2 + 2n + 1) + n^2 = (n^2 + 2nk + k^2 - 2n - 2k + 1) + (n^2 + 2n + 1) \]
This simplifies to:
\[ 2n^2 + 2n + 1 = 2n^2 + 2nk + k^2 - 2k + 1 \]
Cancelling \( 2n^2 + 1 \) from both sides gives:
\[ 2n = 2nk + k^2 - 2k \]
Rearranging yields:
\[ k^2 - 2nk - 2k + 2n = 0 \]

### Step 4: Solve the Quadratic Equation
This is a quadratic equation in \( k \):
\[ k^2 + (2n - 2)k + 0 = 0 \]
Using the quadratic formula:
\[ k = \frac{-(2n - 2) \pm \sqrt{(2n - 2)^2}}{2} \]
This gives:
\[ k = 1 \quad \text{or} \quad k = 0 \]
Since \( k \) must be between \( 0 \) and \( 1 \), we find that \( k \) can take any value in this interval.

### Conclusion
Thus, the solution set of the equation is:
\[ \text{All } x \in \mathbb{R} \text{ such that } x \text{ is an integer.} \]