Example 4 Try to solve the congruence equation
$$x^{2} \equiv -4(\bmod 41)$$
Since 41 is a prime number, it is easy to see that -1 and 4 are quadratic residues modulo 41, so -4 is also a quadratic residue modulo 41, which means the original congruence equation must have a solution.

Because $41=8 \times 5+1$, we have $u=5, \frac{u+1}{2}=3$, and
$$(-4)^{3}=-64 \equiv-23 \equiv 18(\bmod 41)$$

Since
$$(-4)^{5} \equiv(18)(16) \equiv 1(\bmod 41)$$

By (8), the roots we are looking for are
$$x \equiv \pm 18(\bmod 41)$$