Let's find all the square numbers with at most four digits that are equal to one and a half times a cube number!
According to the problem,

$$
x^{2}=1.5 y^{3}
$$

where $x, y$ are integers.

Since $x^{2} \geq 0$ and the number we are looking for is at most a four-digit number, $x^{2} \leq 9999$, and thus $0 \leq y \leq 18$. The $y^{3}$ can only be an even number, otherwise its half would not be an integer. Since $y^{3}$ is even, $x^{2}$ is divisible by 3 (since $2 \cdot 1.5 = 3$), and therefore it is also divisible by 9 (since it is a square number). The $1.5 y^{3}$ can only be divisible by 9 if $y$ is a multiple of 3. From all this, it follows that $y$ is divisible by 6.

Now we only need to examine four cases:

| $y$ | $y^{3}$ | $1.5 y^{3}$ |
| :---: | :---: | :---: |
| 0 | 0 | $\pi$ |
| 6 | 216 | 324 |
| 12 | 1728 | 2592 |
| 18 | 5832 | 8748 |

From the table, we can see that only 0 and 324 are square numbers $\left(0^{2}=1.5 \cdot 0^{3}\right.$ and $\left.18^{2}=1.5 \cdot 6^{3}\right)$, so these are the numbers we are looking for.

Based on the work of Zöldy Balázs (Budapest, Szent István Gymn., I. o.t.)