Example 10 Try to find the solution of the congruence equation
$$x^{2} \equiv 19(\bmod 45)$$
It is obvious that
$x^{2} \equiv 19 \equiv 1(\bmod 9)$ has two roots $x \equiv \pm 1(\bmod 9)$, and $x^{2} \equiv 19 \equiv 4(\bmod 5)$ has two roots $x \equiv \pm 2(\bmod 5)$. From
$$x \equiv a(\bmod 9), x \equiv b(\bmod 5)$$

and the Chinese Remainder Theorem, we get
$$x \equiv 5 \times 2 a+9 \times 4 b=10 a+36 b(\bmod 45)$$

Thus, from
$$x \equiv 1(\bmod 9), x \equiv 2(\bmod 5)$$

we get
$$x \equiv 10+72=82 \equiv-8(\bmod 45)$$

And from
$$x \equiv 1(\bmod 9), x \equiv-2(\bmod 5)$$

we get
$$x \equiv 10-72=-62 \equiv-17(\bmod 45)$$

Therefore, the solutions to the original congruence equation are
$$x \equiv \pm 8, \pm 17(\bmod 45)$$