[Math] Number of solutions to the congruence $x^2\equiv 121\pmod {1800}$

elementary-number-theorymodular arithmeticquadratic-residues

I'm trying to find the number of solutions to this congruence:
$$x^2\equiv 121\pmod {1800}$$

I thought about writing it as a system of congruences. As $1800=3^2 \cdot 5^2 \cdot 2^3$, we get:

$x^2\equiv 121\pmod {5^2} \;,\; x^2\equiv 121\pmod {3^2} \;,\; x^2\equiv 121\pmod {2^3}$

$\Downarrow$

$ x^2\equiv 21\pmod {5^2} \;,\; x^2\equiv 4\pmod {3^2} \;,\; x^2\equiv 1\pmod {2^3} $

Now I'm pretty stuck with how to solve each of the above quadratic congruences, is there a fast way to do it?

Thanks

Hint:

We need $\left(\dfrac x{11}\right)^2\equiv1\pmod{3^2,5^2,2^3}$

Now we can prove $y^2\equiv1\pmod{p^n}$ has exactly two solutions for prime $p\ge3$ and integer $n\ge1$

Finally we can apply Chinese Remainder Theorem to find the number of in-congruent solutions to be $$4\cdot2^2$$