A square number in the decimal system has the form: $\overline{a b 1 a b}$. Which is this square number?
I. Solution. Let the sought number be $2 x$, its square $4 x^{2}$, so according to our data

$$
4 x^{2}=10000 a+1000 b+100+a+b=1000(10 a+b)+100+10 a+b
$$

thus

$$
4 x^{2}-100=1001(10 a+b) \ldots
$$

Since $1001=7 \cdot 11 \cdot 13$

$$
4\left(x^{2}-25\right)=7 \cdot 11 \cdot 13(10 a+b)
$$

Therefore, $10 a+b$ must be a multiple of 4; let's replace it with $4 y$ and since $10 a+b<100$, then $y<25$. This results in

$$
(x+5)(x-5)=7 \cdot 11 \cdot 13 y \ldots
$$

This equation can be interpreted as the factors $7,11,13, y$ need to be grouped into two partial products such that the difference between these two products is 10, while also $y<25$. This cannot be achieved if one or three of the four factors are combined.

We can only meet the requirement if we combine two and two factors, meaning we need to examine the possibilities of the following combinations:

$$
7 \cdot 11 \text { and } 13 y, 7 \cdot 13 \text { and } 11 y, \quad 11 \cdot 13 \text { and } 7 y
$$

I. If one of the partial products is 77, then the other is 67 or 87. Neither of the latter is a multiple of 13 (13y).

II. If one of the partial products is $7 \cdot 13=91$, then the other is 81 or 101. Neither of the latter is a multiple of 11 (11y).

III. If one of the partial products is $11 \cdot 13=143$, then the other is 133 or 153. Of these, only 133 is a multiple of 7:

$$
7 y=133, \quad y=19<25
$$

Accordingly, there is only one solution, and it is:

$$
\begin{gathered}
x^{2}-25=7 \cdot 11 \cdot 13 \cdot 19=19019, \quad x^{2}=19044, \quad x=138 \\
2 x=276 \quad \text { and } \quad 4 x^{2}=276^{2}=76176
\end{gathered}
$$

$4 x^{2}$ is indeed a number of the form $a b 1 a b$!

George Bizám (Bolyai g. VI. o. Bp. V.)

II. Solution. In the square of an even number, the digit in the units place is $0,4,6$. If in the given case $b=0$, then $a$ must also be zero; in this case $00100=10^{2}$.

The square of an even number is a multiple of 4; therefore, $10 a+4$ or $10+6$ must be a multiple of 4, and the following numbers are conceivable:

$24124,44144,64164,84184$,

$16116,36136,56156,76176,96196$

Among these, 44144, 36136, 96196 cannot be square numbers because 441, 361, 961 are square numbers. Among the rest, 76176 is a square number.