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.

Here's a method for avoiding CRT altogether (which comes from thinking about why there can be more than two roots to a quadratic in a commutative ring). If $x$ is any solution, we have
$$
(x - 2)(x + 2) \equiv 0 \bmod 105.
$$
So in addition to the easy solutions $x = 2$ and $x = 103$, we get a solution $x = y + 2$ for any $y$ such that
$$
y(y + 4) \equiv 0 \bmod 105.
$$
I.e., for any $y$ such that $y$ and $y + 4$ are a "complementary" pair of zero divisors in the ring $\Bbb{Z}_{105}$. Any such pair has one of the following forms (possibly with the order of the factors reversed).
$$
\begin{align*}
3m &\cdot 35n & \quad &\mbox{with $1 \le n \le 2$}\\
15m &\cdot 7n & \quad &\mbox{with $1 \le m \le 6$}\\
5m &\cdot 21n & \quad &\mbox{with $1 \le n \le 4$}
\end{align*}
$$
It doesn't take long to work through the $2+6+4$ possibilities for $y$ or $y + 4$ to get that $y(y+4)$ must be one of:
$$
21 \cdot 25\\
35 \cdot 39 \\
45 \cdot 49\\
56 \cdot 60\\
66 \cdot 70\\
80 \cdot 84
$$
Giving 23, 37, 47, 58, 68 and 82 as the not-so-easy solutions.
