Example 3 Try to solve the congruence equation
$$x^{2} \equiv 22(\bmod 29)$$
Solving: Since 29 is a prime number,
$$\begin{aligned}
\left(\frac{22}{29}\right) & =\left(\frac{2}{29}\right)\left(\frac{11}{29}\right)=-\left(\frac{11}{29}\right)=-\left(\frac{29}{11}\right) \\
& =-\left(\frac{7}{11}\right)=\left(\frac{11}{7}\right)=\left(\frac{4}{7}\right)=1
\end{aligned}$$

Therefore, the original congruence equation has a solution.
Since $29=4 \times 7+1$, we have $\frac{u+1}{2}=\frac{7+1}{2}=4$, and
$$22^{4} \equiv(-7)^{4}=49^{2} \equiv 20^{2} \equiv(-9)^{2}=81 \equiv 23(\bmod 29)$$

Because
$$(22)^{7} \equiv(23)(22)(-9) \equiv-1(\bmod 29)$$

and
$$29=3 \times 8+5$$

2 is a quadratic non-residue modulo 29. Since $n=7$, we get
$$b^{n}=2^{7}=2^{5} \times 2^{2}=32 \times 4 \equiv 3 \times 4=12(\bmod 29)$$

From equation (9), we have
$$x \equiv 23 \times 12 \equiv-6 \times 12=-72 \equiv-14(\bmod 29)$$

Therefore, the solution to the original congruence equation is
$$x \equiv \pm 14(\bmod 29)$$