10.21 Find a four-digit number, where the first two digits and the last two digits are the same, and the number itself is equal to the square of an integer.
(Polish Mathematical Olympiad, 1956)
[Solution] Let $x$ be the number we are looking for, then
$$
x=1000 a+100 a+10 b+b
$$

where $a, b$ are integers satisfying $0<a \leqslant 9,0 \leqslant b \leqslant 9$.
Since
$$
x=1100 a+11 b=11(100 a+b) .
$$

$x$ is divisible by 11.
Given that $x$ is a perfect square, $x$ is also divisible by $11^{2}$. Thus we have
$$
x=11(99 a+a+b)=11^{2} \cdot 9 a+11(a+b) .
$$

Therefore, $a+b$ is divisible by 11.
Since
$$
0<a+b \leqslant 18,
$$

we have
$$
a+b=11 \text {. }
$$

Thus
$$
x=11^{2}(9 a+1) \text {. }
$$

Hence, $9 a+1$ is a perfect square.
Let
$$
9 a+1=m^{2} \text {. }
$$

Given $9 a+1 \leqslant 82$, then $m \leqslant 9$.
$$
9 a=(m+1)(m-1) \text {. }
$$

Thus the product $(m+1)(m-1)$ is divisible by 9, but since at most one of the numbers $m+1$ and $m-1$ can be divisible by 3, at most one of them can be divisible by 9. Given $m \leqslant 9$, we can only have $m+1=9$, thus
$$
9 a=(m+1)(m-1)=63 \text {. }
$$

Therefore,
$$
a=7, \quad b=11-a=4 .
$$

Thus, the number that satisfies the condition is
$$
7744=88^{2} .
$$