Example 4 Find all odd prime numbers $p$ for which 11 is a quadratic residue.

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By Theorem 5, we know
$$\left(\frac{11}{p}\right)=(-1)^{(p-1) / 2}\left(\frac{p}{11}\right)$$

By direct calculation, we have
$$\begin{array}{l}
\left(\frac{p}{11}\right)=\left\{\begin{array}{ll}
1, & p \equiv 1,-2,3,4,5(\bmod 11) \\
-1, & p \equiv-1,2,-3,-4,-5(\bmod 11)
\end{array}\right. \\
(-1)^{(p-1) / 2}=\left\{\begin{array}{ll}
1, & p \equiv 1(\bmod 4) \\
-1, & p \equiv-1(\bmod 4)
\end{array}\right.
\end{array}$$

Solving the system of congruences
$$\left\{\begin{array}{l}
x \equiv a_{1}(\bmod 4) \\
x \equiv a_{2}(\bmod 11)
\end{array}\right.$$

we get (left to the reader)
$$x \equiv-11 a_{1}+12 a_{2}(\bmod 44)$$

Combining the above discussions, we know that $\left(\frac{11}{p}\right)=1$ if and only if
$$p \equiv \pm 1, \pm 5, \pm 7, \pm 9, \pm 19(\bmod 44)$$

Therefore, when $p$ is a prime of the above form, 11 is a quadratic residue modulo $p$. Consequently, when $p \neq 11$ is a prime of the following form, $\left(\frac{11}{p}\right)=-1$, i.e., 11 is a quadratic non-residue modulo $p$:
$$p \equiv \pm 3, \pm 13, \pm 15, \pm 17, \pm 21(\bmod 44)$$