21.12. Find a four-digit number that is a perfect square and has the first two digits the same and the last two digits also the same.
21.12. Let $a$ be the first and second digits, $b$ be the third and fourth. Then the given number is equal to $11(b+100 a)$, so $b+100 a=11 x^{2}$ for some natural number $x$. Moreover, $100 \leqslant b+100 a \leqslant$ $\leqslant 908$, hence $4 \leqslant x \leqslant 9$. By calculating the squares of the numbers $44, \ldots, 99$, we find that exactly one of them has the required form: $88^{2}=7744$.