How can I solve the following congruence $x^2 \equiv 9 \pmod {2^3 . 3 . 5^2}$?

Question

How can I solve the following congruence $x^2 \equiv 9 \pmod {2^3 . 3 . 5^2}$?

The problem is that I do not know the number of solutions of $x^2 \equiv 9 \pmod { 3}$, it seems like either it is zero only or any multiple of 3 other than 0, could anyone explain for me why it is not any multiple of 3 other than 0?

Answer 1

$$\begin{align} x^2\equiv 9\bmod 8&\iff x\equiv 1\bmod 2\\ x^2\equiv 9\bmod 3&\iff x\equiv 0\bmod 3\\ x^2\equiv 9\bmod 25&\iff x\equiv \pm 3\bmod 25 \end{align}$$

So, $x^2\equiv 9\mod 600\iff x\equiv \pm3\bmod 150$.

So, the solutions $\pmod{600}$ are going to be $8$, namely $\pm 3\bmod{600}$, $150\pm 3\bmod{600}$, $300\pm3\bmod{600}$, and $450\pm 3\bmod{600}$.