Determine all integers $x$ satisfying
\[ \left[\frac{x}{2}\right] \left[\frac{x}{3}\right] \left[\frac{x}{4}\right] = x^2. \]
($[y]$ is the largest integer which is not larger than $y.$)
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 \( \boxed{0 \text{ and } 24} \).