Find how many solutions the congruence $x^2 \equiv 121 \mod 1800$ has I want to find how many solutions the congruence  $x^2  \equiv 121  \mod 1800$ has.
What is the method to find it without calculating all the solutions? 
I can't use euler criterion here because 1800 is not a primitive root. thanks!!!

Here is a different approach to find the number of solutions to the congruence
$$ x^2 \equiv a \space(\bmod n)$$
If
$$ n = 2^kp_1^{k_1}\cdots p_r^{k_r}$$
where $p_1, \dots ,p_r$ are odd different primes, $k \ge 0$ and $k_1,\dots,k_r \ge 1$.
If the congurence has solutions, the number of the solutions will be equal to $2^r \cdot 2^m,$ Where $$
m= \begin{cases} 
0, & k=0,1\\
1, & k=2\\
2, & k \ge 3
\end{cases} $$
In our case, $$ x^2 \equiv 121 \space(\bmod 1800)$$
$$n = 1800 =2^3\cdot 5^2 \cdot 3^2 \Longrightarrow k=3,r=2$$
And the number of solutions is $2^2 \cdot 2^2 = 16$.
