2. Find all four-digit numbers $m$ for which it is known that:
i) $m$ is a square of a natural number;

ii) the first and second digits of $m$ are the same; and

iii) the third and fourth digits of $m$ are the same.
Solution. Let $x$ be the first, and $y$ be the third digit of $m$ and let $m=k^{2}$. Then

$$
k^{2}=1000 x+100 x+10 y+y=11(100 x+y)
$$

from which it follows that 11 is a divisor of $k$, i.e., $k=11 n$. From

$$
11 \cdot 11 n^{2}=11(100 x+y)
$$

it follows that 11 is a divisor of $x+y$. Since $1 \leq x \leq 9$ and $0 \leq y \leq 9$, we get that $1 \leq x+y \leq 18$, which means that $x+y=11$. Then $n^{2}=9 x+1$, which is possible only for $x=7$. According to this, the desired number $m$ is $7744=88^{2}$.