Graphing Could Fool You

For integer $x$ , $\lceil x^2 \rceil = x^2$ and $\lfloor 2|x| \rfloor = 2|x|$ , therefore,

$\begin{aligned} \lceil x^2 \rceil & = \lfloor 2|x| \rfloor \\ x^2 & = 2|x| = \begin{cases} 2 x & \text{for } x \ge 0 \\ -2x & \text{for } x < 0 \end{cases} \end{aligned}$

$\Rightarrow \begin{cases} x^2 = 2 x & \Rightarrow x(x-2) = 0 & \Rightarrow x = 0, \space 2 & \text{for } x \ge 0 \\ x^2 = -2x & \Rightarrow x(x+2) = 0 & \Rightarrow x = 0, \space -2 & \text{for } x < 0 \end{cases}$

Therefore, there are $\boxed{3}$ integers $-2$ , $0$ and $2$ that satisfy the equation.

Since we are dealing with integers, ceiling and roof functions can be removed. Solve then $x^2=2|x|$

The solution set is $x \in {0, 2, -2}$

Similar solution to @Ahmed R. Maaty

Because we are dealing with integers,floor and ceiling functions can be removed.So what we have is $x^2=2|x|$

Case 1

$\implies \frac{|x|^2}{|x|}=2 \implies |x|=2 \implies x=2,-2$

Case 2

Since it doen't have any addition or subtraction so the answer is also 0

So there are $\boxed{\large{3}}$ solutions, $-2,0,2$