How do you solve $x^2 - 4 \equiv 0 \mod 21$ 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?

You need that $x^2\equiv 4\pmod{3}$ and $x^2\equiv 4\pmod{7}$. For instance, your $x_2=1$ doesn't satisfy $x_2^2\equiv 4\pmod{21}$.
You used or instead of and.
