Find the total number of solutions of [x]^2=x+2{x} where [.] and {.} denotes the greatest integer function and fractional part.

To solve the equation \([x]^2 = x + 2\{x\}\), where \([x]\) is the greatest integer function (also known as the floor function) and \(\{x\}\) is the fractional part of \(x\), we need to break down the components of \(x\) into its integer and fractional parts. Let's denote \([x] = n\) (where \(n\) is an integer) and \(\{x\} = x - n\). This means we can express \(x\) as \(x = n + \{x\}\). Now, substituting \(\{x\}\) into the equation gives us: \[ n^2 = n + 2(x - n) \] This simplifies to: \[ n^2 = n + 2x - 2n \] Rearranging the terms leads to: \[ n^2 + n = 2x \] From this, we can express \(x\) in terms of \(n\): \[ x = \frac{n^2 + n}{2} \] Next, we need to ensure that \(x\) can be expressed as \(n + \{x\}\). Since \(\{x\} = x - n\), we can substitute our expression for \(x\): \[ \{x\} = \frac{n^2 + n}{2} - n = \frac{n^2 - n}{2} \] Now, the fractional part \(\{x\}\) must satisfy \(0 \leq \{x\} < 1\). This gives us the inequality: \[ 0 \leq \frac{n^2 - n}{2} < 1 \] Multiplying through by 2 (which is valid since 2 is positive) results in: \[ 0 \leq n^2 - n < 2 \] This can be factored into: \[ n(n - 1) < 2 \] Now, we can analyze the integer values of \(n\): 1. **For \(n = 0\)**: - \(0(0 - 1) = 0 < 2\) (valid) 2. **For \(n = 1\)**: - \(1(1 - 1) = 0 < 2\) (valid) 3. **For \(n = 2\)**: - \(2(2 - 1) = 2\) (not valid) 4. **For \(n \geq 3\)**: - The product \(n(n - 1)\) will always be greater than or equal to 6, which exceeds 2. Thus, the only valid integer values for \(n\) are 0 and 1. Now, we can find the corresponding \(x\) values for these \(n\): - **For \(n = 0\)**: \[ x = \frac{0^2 + 0}{2} = 0 \] - Here, \(\{x\} = 0\), which is valid. - **For \(n = 1\)**: \[ x = \frac{1^2 + 1}{2} = 1 \] - Here, \(\{x\} = 0\), which is also valid. Now, we check if there are any other solutions by considering the boundaries of \(\{x\}\): - For \(n = 0\), \(x\) can be \(0\) (exactly). - For \(n = 1\), \(x\) can be \(1\) (exactly). Since both cases yield valid solutions, we conclude that the total number of solutions to the equation \([x]^2 = x + 2\{x\}\) is:

• 1 solution for \(n = 0\) (i.e., \(x = 0\))
• 1 solution for \(n = 1\) (i.e., \(x = 1\))