There is an example in my textbook of how you solve:
x^2 -4\equiv 0 \mod 21 \Leftrightarrow x^2-4\equiv 0 \mod 3 \times 7
and then 2 congruences can be formed out of this equation if:
x^2-4\equiv0 \mod 3 \\ x^2-4 \equiv 0 \mod 7
and from these 2 congruences result 2 more congruences, for each:
x - 2 \equiv 0 \mod 3 \Rightarrow x_1 = 2\\ x + 2 \equiv 0 \mod 3 \Rightarrow x_2 = 1\\ x - 2 \equiv 0 \mod 7 \Rightarrow x_3 = 2\\ x + 2\equiv 0 \mod 7 \Rightarrow x_4 = 5
and then 4 systems of linear congruences are formed:
\begin{cases} x \equiv 2 \mod 3 \\ x \equiv 2 \mod 7\end{cases}
\begin{cases} x \equiv 2 \mod 3 \\ x \equiv 5 \mod 7\end{cases}
\begin{cases} x \equiv 1 \mod 3 \\ x \equiv 2 \mod 7\end{cases}
\begin{cases} x \equiv 1 \mod 3 \\ x \equiv 5 \mod 7\end{cases}
What is the purpose of these systems? I already have the 4 solutions (x_1, x_2, x_3, x_4) of the congruence. Why do I need to form these systems?
Answer
Your ‘4 solutions’ are not solutions modulo 21, but pairs of solutions \bmod3 on one hand, \bmod 7 on the other hand.
From these pairs of solutions, you recover solutions modulo 3\times 7 with the Chinese remainder theorem.
Start from the Bézout's relation \;5\cdot 3-2\cdot 7=1. Then the solution corresponding to the pair (\color{red}1\bmod3,\color{red}5\bmod7), for instance, will be
x\equiv\color{red}5\cdot5\cdot 3-\color{red}1\cdot2\cdot 7=61\equiv19\mod 21.
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