$\overline{a b}$, a two-digit number written in the decimal system, its square is a three-digit number: $\overline{a b^{2}}=\overline{c d e}$.

The square of the two-digit number obtained by reversing the digits is also a three-digit number, and this latter square number is also obtained by reversing the order of the digits of the following square number: $\overline{b a^{2}}=\overline{e d c}$.

How many such two-digit numbers are there and which are they?
The numbers $\overline{a b^{2}}$ and $\overline{b a^{2}}$ are three-digit numbers, where the first digit can only be a number whose square is a single digit. Therefore, both $a$ and $b$ can only be $1, 2, 3$.

(We must exclude the cases where $a=0$ or $b=0$.)

Thus, the second digit of the numbers $\overline{a b}$ and $\overline{b a}$ must also have a square that is a single digit, which implies that twice the product of $a$ and $b$ must also be a single digit, i.e.,

$$
2 a b < 10, \quad a b < 5
$$

and these conditions, if $0 < a \leq 3$ and $0 < b \leq 3$, are only satisfied by the pairs

$$
(1,1), \quad(1,2), \quad(1,3), \quad(2,2)
$$

Therefore, the solutions to the problem are:

$$
11,12,13,21,22,31
$$

Indeed: $11^{2}=121, 12^{2}=144$ and $21^{2}=441, 13^{2}=169, 31^{2}=961, 22^{2}=484$.

George Barna (Gr. Klebersberg Kunó g. VI. o., Hatvan)