1. Determine all distinct pairs of natural numbers $(x, y)$ such that by swapping the last two digits of the number $x^{2}$, the number $y^{2}$ is obtained.
Solution. Let's note that if the pair $(x, y)$ satisfies the condition of the problem, then the pair $(y, x)$ also satisfies the condition of the problem. Therefore, it is sufficient to determine the pairs $(x, y)$ for which $x>y$.

Let

$$
x^{2}=\overline{a_{1} a_{2} \ldots a_{k} p q} \text { and } y^{2}=\overline{a_{1} a_{2} \ldots a_{k} q p} .
$$

Then

$$
x^{2}-y^{2}=9(p-q)
$$

We will prove that $d=p-q$ cannot be any of the numbers $9,8,7$ and 6.

a) If $d=9$, then $p=9$ and $q=0$, which is not possible because the number $x^{2}$ cannot have a two-digit ending 90.

b) If $d=8$, then $(p, q)=(9,1)$ or $(p, q)=(8,0)$. This is also not possible, because the two-digit ending of $x^{2}$ cannot be either 91 or 80.

c) If $d=7$, then $(p, q) \in\{(7,0),(8,1),(9,2)\}$, which is again not possible because the two-digit ending of $x^{2}$ cannot be either 92, 81, or 70.

d) If $d=6$, then $(p, q) \in\{(6,0),(7,1),(8,2),(9,3)\}$, which for the same reasons as above is not possible.

Thus, $d \leq 5$. From this it follows that

$$
2 y+1 \leq x+y \leq(x-y)(x+y)=9 d \leq 45
$$

from which it follows $y \leq 22$. By checking all values of $y$ from 1 to 22, it is easily verified that $y=13$ or $y=14$, i.e., the desired pairs are $(13,14)$ and $(14,13)$.