Let's determine that two-digit number, the square of which is a four-digit number where the first and second digits, as well as the third and fourth digits, are identical.
I. solution: Let the sought number be $10a + b$, then according to the problem,

$$
(10a + b)^2 = 1000c + 100c + 10d + d = 11(100c + d)
$$

where $a, b, c, d$ are non-negative single-digit numbers, and $a \neq 0, c \neq 0$.

Since the right-hand side is divisible by 11, the left-hand side must also be divisible by 11. Since 11 is a prime number, $(10a + b)^2$ can only be divisible by 11 if $10a + b$ itself is divisible by 11.

$10a + b$ can only be divisible by 11 if $a = b$, that is, $10a + b = 11a$, so (1) can be written as

$$
11^2 a^2 = 11(100c + d)
$$

which simplifies to

$$
11a^2 = 100c + d
$$

from which

$$
a^2 = \frac{100c + d}{11} = 9c + \frac{c + d}{11}
$$

where $\frac{c + d}{11}$ must be an integer. Since $c = d = 0$, and $c = d = 9$ are not possible (9999 is not a perfect square), we have $0 < c + d < 18$, and thus $c + d = 11$. Therefore,

$$
a^2 = 9c + 1
$$

from which

$$
c = \frac{a^2 - 1}{9} = \frac{(a + 1)(a - 1)}{9}
$$

Neither $(a + 1)$ nor $(a - 1)$ can be divisible by 3, so either $(a + 1)$ or $(a - 1)$ must be divisible by 9.

$a - 1 \neq 0$, otherwise $a = 1$ and $c = 0$, which is impossible. Since $a \leq 9$, $a - 1 \neq 9$. Therefore, it must be

$$
a + 1 = 9, \quad \text{hence} \quad a = 8
$$

Thus, the sought number is $11a = 88$. Indeed, $88^2 = 7744$.

Kálmán Györy (Ózd, József Attila g. II. o. t.)

II. solution: According to the first solution, the sought number is divisible by 11, and its square is a four-digit number, so only $33, 44, \ldots, 99$ are considered.

Among these, the odd numbers do not meet the criteria, because

$$
(10a + a)^2 = 100a^2 + 20a^2 + a^2
$$

and if $a$ were odd, then $a^2$ would have an odd number in the units place (09, 25, 49, 81), and the tens place would also be even due to $20a^2$ - an even number would stand, so the third and fourth digits could not match.

Therefore, only 44, 66, or 88 can be considered. Squaring each of them, we can see that only 88 satisfies the problem's requirements.

László Kovács (Bp., V., Eötvös g. III. o. t.)