AustriaMO

Since $x^2 \ge 0$ certainly holds, we must have $x \ge 0$. $x = 0$ is obviously a solution. We now assume $x > 0$. Since $\lfloor y \rfloor \le y$, we certainly have
$$x^2 \le \frac{x}{2} \cdot \frac{x}{3} \cdot \frac{x}{4} = \frac{x^3}{24} \Rightarrow 24 \le x.$$
Furthermore, since $\left\lfloor \frac{x}{2} \right\rfloor \ge \frac{x}{2} - \frac{1}{2}$, $\left\lfloor \frac{x}{3} \right\rfloor \ge \frac{x}{3} - \frac{2}{3}$ and $\left\lfloor \frac{x}{4} \right\rfloor \ge \frac{x}{4} - \frac{3}{4}$, hold, we also have
$$\left(\frac{x}{2} - \frac{1}{2}\right) \left(\frac{x}{3} - \frac{2}{3}\right) \left(\frac{x}{4} - \frac{3}{4}\right) = \frac{1}{24}(x-1)(x-2)(x-3) \le \left\lfloor \frac{x}{2} \right\rfloor \cdot \left\lfloor \frac{x}{3} \right\rfloor \cdot \left\lfloor \frac{x}{4} \right\rfloor = x^2,$$
which is equivalent to $x^3 - 6x^2 + 11x - 6 \le 24x^2$ or $x^3 - 30x^2 + 11x - 6 \le 0$. This can only hold for $x < 30$, since both $x^3 \ge 30x^2$ and $11x > 6$ hold for $x \ge 30$. We see that further solutions $x$ can only exist for $24 \le x \le 29$.

For $x = 24$, we have
$$\left[ \frac{24}{2} \right] \cdot \left[ \frac{24}{3} \right] \cdot \left[ \frac{24}{4} \right] = 12 \cdot 8 \cdot 6 = 24^2,$$
and this is a solution. For $x = 25/26$, $x = 27$ and $x = 28$, the expression $\left[\frac{x}{2}\right] \cdot \left[\frac{x}{3}\right] \cdot \left[\frac{x}{4}\right]$ yields $13 \cdot 8 \cdot 6$, $13 \cdot 9 \cdot 6$ and $14 \cdot 9 \cdot 7$ respectively, none of which is a perfect square.
We see that the integer solutions of the equation are exactly the numbers 0 and 24. $\square$