Problem 1356

To determine all integers $x$ satisfying the equation
$$\left\lfloor \frac{x}{2} \right\rfloor \left\lfloor \frac{x}{3} \right\rfloor \left\lfloor \frac{x}{4} \right\rfloor = x^2,$$
we will follow these steps:

1. **Non-negativity of $x$**:
Since $x^2$ is nonnegative, $x$ must also be nonnegative. Therefore, we only need to consider $x \geq 0$.

2. Bounding the product:
We need to analyze the product $\left\lfloor \frac{x}{2} \right\rfloor \left\lfloor \frac{x}{3} \right\rfloor \left\lfloor \frac{x}{4} \right\rfloor$. For large $x$, this product can be approximated by:
$$\left( \frac{x}{2} - 1 \right) \left( \frac{x}{3} - 1 \right) \left( \frac{x}{4} - 1 \right).$$
Simplifying this approximation:
$$\left( \frac{x}{2} - 1 \right) \left( \frac{x}{3} - 1 \right) \left( \frac{x}{4} - 1 \right) = \frac{x}{2} \cdot \frac{x}{3} \cdot \frac{x}{4} - \text{lower order terms}.$$
$$= \frac{x^3}{24} - \text{lower order terms}.$$
For large $x$, $\frac{x^3}{24}$ grows faster than $x^2$, so the product will exceed $x^2$. Therefore, we need to check $x$ for smaller values.

3. **Checking values of $x$ from 0 to 29**:
We need to check each integer $x$ from 0 to 29 to see if it satisfies the equation. We will compute the product and compare it to $x^2$.

- For $x = 0$:
$$\left\lfloor \frac{0}{2} \right\rfloor \left\lfloor \frac{0}{3} \right\rfloor \left\lfloor \frac{0}{4} \right\rfloor = 0 \cdot 0 \cdot 0 = 0,$$
which equals $0^2 = 0$. So, $x = 0$ is a solution.

- For $x = 24$:
$$\left\lfloor \frac{24}{2} \right\rfloor \left\lfloor \frac{24}{3} \right\rfloor \left\lfloor \frac{24}{4} \right\rfloor = 12 \cdot 8 \cdot 6 = 576,$$
which equals $24^2 = 576$. So, $x = 24$ is a solution.

- For other values of $x$ from 1 to 23 and 25 to 29, we need to check each one individually. For example:
- For $x = 1$:
$$\left\lfloor \frac{1}{2} \right\rfloor \left\lfloor \frac{1}{3} \right\rfloor \left\lfloor \frac{1}{4} \right\rfloor = 0 \cdot 0 \cdot 0 = 0,$$
which does not equal $1^2 = 1$. So, $x = 1$ is not a solution.
- For $x = 2$:
$$\left\lfloor \frac{2}{2} \right\rfloor \left\lfloor \frac{2}{3} \right\rfloor \left\lfloor \frac{2}{4} \right\rfloor = 1 \cdot 0 \cdot 0 = 0,$$
which does not equal $2^2 = 4$. So, $x = 2$ is not a solution.
- Continue this process for all $x$ from 1 to 23 and 25 to 29.

4. Conclusion:
After checking all values, we find that the only solutions are $x = 0$ and $x = 24$.

The final answer is $\boxed{0 \text{ and } 24}$.