[ Decimal number system ] $[$ Equations in integers $]$

Find a four-digit number that is a perfect square and such that the first two digits are the same as each other and the last two digits are also the same.
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, $100 \leq b+100 a \leq 908$, hence $3 \leq x \leq 9$.

Calculating the squares of the numbers $33,44, \ldots, 99$, we find that exactly one of them has the required form: $88^{2}=7744$.

## Answer

7744.