P r o b l e m 6. Let's find a four-digit number that is a perfect square, if its first two digits and last two digits are equal.
Solution. Let $N=\overline{x x y y}$ be the desired number. Write it as a systematic number in base 10 and simplify:

$$
\begin{aligned}
& N=1000 x+100 x+10 y+y \\
& N=1100 x+11 y \\
& N=11(100 x+y)
\end{aligned}
$$

Since (by condition) $N$ is the square of some natural number, the second factor $100 x+y$ must be divisible by 11, i.e., $100 x+y \vdots 11$, or $x+y \vdots 11$ (since $99 x \vdots 11$).

Since $0 \leqslant x \leqslant 9$ and $0 \leqslant y \leqslant 9$, we get the only option: $x+y=11$. Hence, $y=11-x$, and therefore

$$
100 x+(11-x)=11 t^{2}, \text { or } 9 x+1=t^{2}
$$

By trying all values of $x$ from 1 to 9, we find the desired number $N=7744=88^{2}$.