[Math] Solve $x^2=b \mod m$ congruence equations

Question

I would like to solve $x^2=b \mod m$ type of equations.
I have, for example, have to find all $a$ for which $a^2= 3 \mod 11$ and I have a couple of very basic (I'm sorry!) questions

1. How do I use quadratic reciprocity to prove that this has a
   solution?

2. I can see that 5 and 6 are solutions and I think that it suffices to
   try all numbers from 0 to 10 but I do not understand why (can I try
   even less numbers?). If $a$ was not squared I clearly see that the
   numbers from 0 to 10 generate all congruence classes but with the
   square there I do not get all congruence classes for $\pmod {11}$.

3. Is there a general way to solve this if I have to try more than 11
   numbers?

Sorry for my basic questions and thanks in advance for the answer!

Answer 1

Using quadratic reciporcity:

$$\left(\frac{3}{11}\right)=-\left(\frac{11}{3}\right)=-\left(\frac{2}{3}\right)=-(-1)=1$$

You can always try only with the elements

$$\left\{0,1,2,\ldots,\frac{p-1}{2}\right\}$$

as the other elements are negatives of some of the above ones and thus squared...