elementary number theory - Solving the congruence $x^2 equiv 4 mod 105$. Is there an alternative to using Chinese Remainder Theorem multiple times?

I'm trying to solve $$x^2 \equiv 4 \mod 105.$$

This is of course equivalent to $$(x+2)(x-2) \equiv 0 \mod 105$$

which is also equivalent to the system of congruences

$$(x+2)(x-2) \equiv 0 \mod 3$$
$$(x+2)(x-2) \equiv 0 \mod 5$$
$$(x+2)(x-2) \equiv 0 \mod 7$$

which have solutions of $$x \equiv 1\ \text{or}\ 2$$
$$x \equiv 2\ \text{or}\ 3,\ \text{and}$$

$$x \equiv 2\ \text{or}\ 5$$
respectively.

Now, I could in principle take each combination of {$1,2$}, {$2,3$}, and {$2,5$} and use the Chinese Remainder Theorem to solve each system, but that seems incredibly tedious.

Is there a simpler way?

I'll note that this is a homework question, so it seems likely that there's a trick. Unfortunately I can't spot one.