A six-digit square number is divided into three two-digit numbers. The two outer two-digit numbers are equal, and the middle one is half of one of these. Which is this six-digit square number?
Let the square number in question be $n^{2}$ and denote the number formed by its middle two digits by $x$. Thus, the two outer two-digit numbers are $2 x$, so

$$
10 \leqq 2 x < 100
$$

and thus

$$
5 \leqq x < 50
$$

Furthermore,

$$
n^{2}=2 x \cdot 10^{4} + x \cdot 10^{2} + 2 x = 20102 x = 2 \cdot 19 \cdot 23^{2} \cdot x
$$

A square number, when expressed as a product of powers of different prime numbers, has all exponents even. Therefore, in such an expression for $x$, the prime numbers 2 and 19 appear with odd exponents, so their exponents must be at least 1, while the exponents of all other prime powers are even. Thus, by isolating the first powers of 2 and 19 and combining the rest, $x$ can be written as:

$$
x=2 \cdot 19 \cdot k^{2}=38 k^{2}, \quad \text{and thus} \quad n^{2}=(2 \cdot 19 \cdot 23 \cdot k)^{2}
$$

where $k$ is an integer. (1) is only satisfied when $k=1$, so the only possible solution is:

$$
n^{2}=(2 \cdot 19 \cdot 23)^{2}=874^{2}=763876
$$

This indeed meets the requirements of the problem.