Example 4 Find an odd prime $p$ such that
$$x^{2}+3 \equiv 0(\bmod p)$$

has a solution.
From $p$ being a prime number and Lemma 8 and Theorem 2, we have
$$\begin{array}{l}
\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right) \\
\quad=(-1)^{\frac{p-1}{2}}(-1)^{\frac{p-1}{2}}\left(\frac{p}{3}\right)=\left(\frac{p}{3}\right) .
\end{array}$$

Thus, when $p \equiv 1(\bmod 3)$, we have $\left(\frac{-3}{p}\right)=\left(\frac{1}{3}\right)=1$, meaning that equation (45) has a solution. When $p \equiv -1(\bmod 3)$, we have $\left(\frac{-3}{p}\right)=\left(\frac{-1}{3}\right)=(-1)^{\frac{3-1}{2}}=-1$, meaning that equation (45) has no solution. Additionally, for $p=3$, equation (45) clearly also has a solution.