Solve the equation $ \lceil x^2 - x \rceil = x + 3 $ for $x \in \mathbb{R}$, where $ \lceil x \rceil $ denotes the ceiling function.

1. Problem Description

Solve the equation $\lceil x^2 - x \rceil = x + 3$ for $x \in \mathbb{R}$, where $\lceil x \rceil$ denotes the ceiling function.

2. Solution Steps

Since $\lceil x^2 - x \rceil = x + 3$, and the left side is an integer, $x + 3$ must be an integer. Therefore, $x$ must be an integer. Let $x = n$, where $n$ is an integer. Then the equation becomes

$\lceil n^2 - n \rceil = n + 3$

Since $n$ is an integer, $n^2 - n$ is also an integer, so $\lceil n^2 - n \rceil = n^2 - n$.

The equation is now $n^2 - n = n + 3$, or $n^2 - 2n - 3 = 0$.

This is a quadratic equation in $n$. Factoring gives $(n - 3)(n + 1) = 0$.

So $n = 3$ or $n = -1$.

Now we check if these solutions are valid.

If $x = 3$, then $\lceil 3^2 - 3 \rceil = \lceil 9 - 3 \rceil = \lceil 6 \rceil = 6$.

Also, $x + 3 = 3 + 3 = 6$. So $x = 3$ is a solution.

If $x = -1$, then $\lceil (-1)^2 - (-1) \rceil = \lceil 1 + 1 \rceil = \lceil 2 \rceil = 2$.

Also, $x + 3 = -1 + 3 = 2$. So $x = -1$ is a solution.

Now let's consider the general case where $x$ is not an integer. Let $x = n + \alpha$, where $n$ is an integer and $0 < \alpha < 1$. Then we have

$\lceil (n + \alpha)^2 - (n + \alpha) \rceil = n + \alpha + 3$

$\lceil n^2 + 2n\alpha + \alpha^2 - n - \alpha \rceil = n + \alpha + 3$

Since the ceiling function gives an integer value, $n + \alpha + 3$ must be an integer. Then $\alpha$ must be an integer. But since $0 < \alpha < 1$, $\alpha$ cannot be an integer. Thus, $x$ must be an integer.

3. Final Answer

$x = 3$ or $x = -1$