1. Determine a four-digit number that is a square of a natural number and where the first two digits and the last two digits are equal.
Solution. From the condition of the problem, it follows that

$$
\overline{a a b b}=1100 a+11 b=11(100 a+b)=k^{2}, \quad k \in \mathbb{Z}
$$

which means that $100 a+b=11 l^{2}, l \in \mathbb{Z}$, where $a, b \in\{1,2, \ldots, 9\}$. By direct verification, the only possibility is $l=8, a=7, b=4$.

Indeed, $88^{2}=7744$.