Find those numbers which, when their decimal representation is appropriately split into two parts and these parts are considered as independent numbers, the product of these two parts equals half of the original number (e.g., $1352=2 \cdot 13 \cdot 52$). Which of these are perfect squares?
When the decimal representation of the number $N$ is split into two parts, the number represented by the sequence of digits before the split is $P$, the number represented by the sequence of digits after the split is $Q$, and the number of digits after the split is $q$. Then $N$ can be written in two ways:

$$
P \cdot 10^{q}+Q=2 \cdot P \cdot Q
$$

or, rearranged,

$$
(2 P-1) Q=P \cdot 10^{q}
$$

According to this, $2 P-1>1$, because $Q>1$.

The number $N$ we are looking for is

$$
\begin{gathered}
N=2 \cdot P \cdot Q=2^{q-1} \cdot 5^{q-r} \cdot\left(5^{r}+1\right)^{2} \\
{\left[=2^{q} \cdot 5^{q-r} \cdot \frac{5^{r}+1}{2} \cdot 5^{r}+2^{q-1} \cdot 5^{q-r}\left(5^{r}+1\right)=10^{q} \cdot P+Q\right]}
\end{gathered}
$$

and this is a perfect square if and only if $q-1$ and $q-r$ are even, that is, if $q$ and $r$ are both odd.

If $r<q$, then $Q$ ends with $q-r$ zeros. It is easy to see that by omitting these, applying the split at the same point, the resulting number $N^{\prime}$ also satisfies the conditions, so it is enough to consider the solutions corresponding to the case $r=q$. For these, when $q=1,2,3,4$, we get respectively

$$
N=3\left|6\left(=6^{2}\right), \quad 13\right| 52, \quad 63\left|504\left(=252^{2}\right), \quad 313\right| 5008
$$

József Váradi (Budapest, Ságvári E. Gymnasium, 4th grade) György Szabó (Nyíregyháza, Vasvári P. Gymnasium, 3rd grade)