Number of solutions to the congruence $x^2\equiv 121\pmod {1800}$ 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 $$4\cdot2^2$$
See also: Number of solutions of $x^2=1$ in $\mathbb{Z}/n\mathbb{Z}$
