*8. Find all natural numbers $x$, such that $x^{2}$ \% is a 12-digit number of the form $x^{2}=2525 * * * * * * 9$ (the six unknown digits represented by $*$ do not necessarily have to be the same).





The translation is provided as requested, preserving the original text's line breaks and format.
The last two digits of the required number $x$ can be 17, 33, 67, 83 (see Question 8 of Grade 8). To ensure that the first four digits of the square of the number are 2525, the inequality must be formed as follows:
$$
2525 \cdot 10^{8}<x^{2}<2526 \cdot 10^{8} \text {. }
$$

This simplifies to
$$
\sqrt{2525} \cdot 10^{4}<x<\sqrt{2526} \cdot 10^{4}.
$$
Taking the square root, we get
$$
\begin{array}{l}
\sqrt{2525}=50.2493 \cdots, \\
\sqrt{2526}=50.2593 \cdots .
\end{array}
$$

Therefore, $502493<x<502594$. Within this interval, the numbers whose last two digits are 17, 33, 67, 83 are four, which are: $502517, 502533, 592567, 502583$.