<s>[INST] How many solutions does $x^2 \equiv {-1} \pmod {365}$ have? 
How many solutions does $x^2 \equiv {-1} \pmod {365}$ have?

My thought:
$365 = 5 \times 73$ where $5$ and $73$ are prime numbers. 
So we can obtain $x^2 \equiv {-1} \pmod 5$ and $x^2 \equiv {-1} \pmod {73}$.
For $x^2 \equiv {-1} \pmod 5$, 
we checked $5 \equiv 1 \pmod 4$, therefore it is solvable. 
Using the Euler criterion, $(-1)^{\frac{5-1}{2}} \equiv 1 \pmod 5$. 
Thus it has two solutions.
For $x^2 \equiv {-1} \pmod {73}$, 
we checked $73 \equiv 1 \pmod 4$, therefore it is solvable. 
Using the Euler criterion, $(-1)^{\frac{73-1}{2}} \equiv 1 \pmod{73}$. 
Thus it has two solutions.
So, there are four solutions in total.
 [/INST] Working modulo $\,73\,$ :
$$2^6=-9\implies 2^{12}=(-9)^2=8=2^3\implies 2^9=1$$
$$3^4=8=2^3\implies3^{12}=(2^3)^3=2^9=1$$
$$5^6=3\implies 5^{36}=3^6=8\cdot3^2=72=-1$$
Thus, we've found a primitive root modulo $\,73\,$ (namely $\,5\,$), and from here we get two solutions to $\,x^2=-1\pmod{73}\,$ : $\;\;5^{18}=(5^6)^3=3^3=27\;,\;-27=46\;$ , and the two solutions to $\,x^2=-1\pmod 5 \;$ are $\,\;2,3\;$
So we have the solutions modulo $\,365\,$ : $\;173\;,\;-173=192\;,\;338\ldots$ and I think that's all, folks: four different solutions: $\;27,\; 46,\; 173,\;  339\pmod{365}$
</s>