AustriaMO2013

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 $\lfloor \frac{x}{2}\rfloor \ge \frac{x}{2}-\frac{1}{2}$, $\lfloor \frac{x}{3}\rfloor \ge \frac{x}{3}-\frac{2}{3}$ and $\lfloor \frac{x}{4}\rfloor \ge \frac{x}{4}-\frac{3}{4}$, hold, we also have $$\left(\frac{x}{2}-\frac{1}{2}\right)\left(\frac{x}{3}-\frac{2}{3}\right)\left(\frac{x}{4}-\frac{3}{4}\right)=\frac{1}{24}(x-1)(x-2)(x-3)\le \lfloor \frac{x}{2}\rfloor \cdot \lfloor \frac{x}{3}\rfloor \cdot \lfloor \frac{x}{4}\rfloor =x^2,$$ 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. $\square$