When the last two digits of the square of a natural number are swapped, we get the square of the number that is one greater. Determine all such numbers.

Határozzuk meg az összes ilyen számot.

(Note: The last sentence is kept in Hungarian as it was in the original text, as it seems to be a repetition or a specific instruction in the original context.)
Let the last two digits of $n^{2}$ be $A$ and $B$ in this order. Then the last two digits of $(n+1)^{2}$ are $B$ and $A$. The difference between the two numbers is:

$$
(n+1)^{2}-n^{2}=2 n+1=(10 B+A)-(10 A+B)=9(B-A)
$$

From this, it immediately follows that $B-A$ is odd. Neither $A$ nor $B$ can be 0, because a square number ending in 0 has its last two digits as 0. This also means that $B>A$, and since $B \leq 9, B-A \leq 8$ (and odd), the possible values for $B-A$ are: $1, 3, 5, 7$. From this, we get the following values for $n$: $4, 13, 22, 31$.

The squares of the corresponding pairs are:

$$
16,25 ; \quad 169,196 ; \quad 484,529 ; \quad 961,1024
$$

Among these, only the pair 169, 196 meets the requirements of the problem, so the only solution is $n=13$.