A four-digit number, which is a perfect square of a number, has its first two and last two digits equal; find this number.
The sought number has the following form:

$$
\begin{gathered}
a^{2}=1000 x+100 x+10 y+y= \\
=100 x(10+1)+y(10+1)=11(100 x+y)
\end{gathered}
$$

Since $11(100 x+y)$ is a perfect square, it is necessary that $100 x+y=99 x+x+y$ be divisible by 11; that is, $x+y$ must be a multiple of 11. But since both $x$ and $y$ are less than 10, it follows that

$$
x+y=11
$$

from which

$$
y=11-x
$$

Substituting this into $100 x+y$, we get

$$
100 x+y=100 x+11-x=99 x+11=11(9 x+1)
$$

Therefore,

$$
a^{2}=11^{2}(9 x+1)
$$

Since $(9 x+1)$ is a perfect square, the last digit of $(9 x+1)$ can only be 1, 4, 5, 6, or 9. Thus, the last digit of the product $9 x$ can only be 0, 3, 4, 5, or 8; hence, the value of $x$ can only be 2, 5, 6, or 7. If $x$ is equal to 2, 5, or 6, then $9 x+1$ is equal to 19, 46, or 55, none of which are perfect squares. But if $x=7$, then $9 x+1=64=8^{2}$. Therefore, only this value satisfies the problem. If $x=7$, then $y=4$ and thus the sought number is 7744.

The problem was solved by: Frankl I., Krisztián Gy., Prohászka J., Weisz J.