9. Solve the congruence equation $x^{2} \equiv 59(\bmod 125)$
9. Solution: First, it is easy to see that $59 \equiv 9\left(\bmod 5^{2}\right)$, hence
$$x^{2} \equiv 59\left(\bmod 5^{2}\right)$$

has solutions $x \equiv \pm 3(\bmod 25)$. Next, we solve
$$x^{2} \equiv 59\left(\bmod 5^{3}\right)$$
(1) Let $x=25 t+3$, substituting we get
$$(25 t+3)^{2} \equiv 59\left(\bmod 5^{3}\right)$$

Thus,
$$(6)(25) t \equiv 50\left(\bmod 5^{3}\right)$$

Dividing both sides by 5, we get
$$6 t \equiv 2(\bmod 5)$$

This simplifies to
$$t \equiv 2(\bmod 5)$$

Substituting $x=25 t+3$ gives one solution as $x=25(5 k+2)+3=53\left(\bmod 5^{3}\right)$.
(2) Let $x=25 t-3$, substituting we get
$$(25 t-3)^{2} \equiv 59\left(\bmod 5^{3}\right)$$

Expanding, we get
$$(-6)(25) t \equiv 50\left(\bmod 5^{3}\right)$$

Dividing by 25, we get
$$-6 t \equiv 2(\bmod 5)$$

This simplifies to
$$t \equiv -2(\bmod 5)$$

Thus, the second solution is $x_{2}=25(5 k-2)-3 \equiv -53\left(\bmod 5^{3}\right)$.
In summary, the solutions are $x \equiv \pm 53\left(\bmod 5^{3}\right)$.
Note: Regarding the solution of higher-order congruence equations of the form
$$f(x) \equiv 0\left(\bmod p^{\alpha}\right)$$
where $f(x)$ is an $n$-degree polynomial with integer coefficients, $p$ is a prime, and $x \geqslant 1$, it is closely related to the corresponding congruence equation
$$f(x) \equiv 0(\bmod p)$$

Due to the depth of the involved knowledge, it will not be elaborated here. Readers interested in this topic can refer to Professor Hua Luogeng's "Introduction to Number Theory" and other specialized works.