Example 6 Discuss the congruence equation
$$x^{2} \equiv -286 \pmod{4272943}$$

whether it has a solution, where 4272943 is a prime number.
Let $p=4272943$, by Lemma 7 we have
$$\left(\frac{-286}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right)\left(\frac{143}{p}\right)$$

Since $4272943 \equiv 7(\bmod 8)$, we have
$$\left(\frac{-1}{p}\right)=-1,\left(\frac{2}{p}\right)=1$$

Thus,
$$\left(\frac{-286}{p}\right)=-\left(\frac{143}{p}\right)$$

Since $143=4 \times 35+3, p=3(\bmod 4)$, by Theorem 3 we have
$$\left(\frac{143}{p}\right)=-\left(\frac{p}{143}\right)$$

From $p=143 \times 29880+103$, we get
$$\left(\frac{p}{143}\right)=\left(\frac{103}{143}\right)$$

By Theorem 3 and $103=3(\bmod 4), 143 \equiv 3(\bmod 4)$, we have
$$\begin{array}{l}
\left(\frac{103}{143}\right)=-\left(\frac{143}{103}\right)=-\left(\frac{40}{103}\right)=-\left(\frac{2^{2} \times 2 \times 5}{103}\right) \\
\quad=-\left(\frac{2 \times 5}{103}\right)=-\left(\frac{2}{103}\right)\left(\frac{5}{103}\right)=-\left(\frac{5}{103}\right) \\
=-\left(\frac{103}{5}\right)=-\left(\frac{3}{5}\right)=1
\end{array}$$

Therefore,
$$\left(\frac{-286}{p}\right)=1$$

This means that equation (51) has a solution.