Which numbers have the property that their last two digits are the same as the last two digits of their square?
The last few digits of a number's square are only influenced by the same number of digits at the end of the base, because $\left(10^{k} a+b\right)^{2}=10^{k}\left(10^{k} a^{2}+2 a b\right)+b^{2}$. Thus, the last digit of the sought numbers can only be $0,1,5$ or 6, because their squares also end in the same digit.

Now let's find the second last digit. If the last digit is $0, (10 x+0)^{2}=100 x^{2}$, so the second last digit can only be $x=0$. $(10 y+1)^{2}=100 y^{2}+20 y+1$, so $2 y=y, y=0$ must be. If the last digit is $5, (10 z+5)^{2}-100\left(z^{2}+z\right)+25$, so $z=2$. Finally, if the last digit is $6, (10 u+6)^{2}=100\left(u^{2}+u\right)+10(2 u+3)+6$, so $2 u$ plus 3 can only differ from $u$ by a multiple of 10, in other words $10 \mid(u+3), u=7$.

Thus, the squares of numbers ending in $00,01,25,76$ end in the same two digits.

Note: The squares of decimal fractions ending in 01, 25, 76 also end in these digits. (The 00 cannot be considered a "termination" of a decimal fraction.) Indeed, GacsÁLYI SÁNDOR correctly notes that not only the second, but all higher integer powers of these numbers also end in the same digits as the base. (1 point) GEHÉr LáSZLÓ submitted a solution that does not break down the sought number into digits. ( $2 \times 3$ points.) Try to find this solution as well!