Solutions of $y^2 = \alpha$ in $\mathbb{F}_{19}$ So I'm working on an exercise for elliptic curves and in one of my steps I have to determine all numbers $y \in \mathbb{F}_{19}$ for which it holds that $y^2 = \alpha$, with $\alpha \in \mathbb{F}_{19}$. Now, I could check every possibility by hand, but that seems a bit excessive. Are there any methods to solve this kind of problem? 
As always, any help would be greatly appreciated.
 A: $y^2\equiv \alpha\iff y\equiv \pm k$, where $k^2\equiv \alpha\pmod{\! 19}$. First, you see if such $k$ exists by using Quadratic reciprocity. If it exists, to find the $k$, you need to check $\{0,1,2,\ldots,9\}$, and there is no simple algorithm that wouldn't use trial-and-error (there is Cipolla's algorithm, but it needs some trial-and-error too).
A: The order of the field is small enough that brute force---that is, simply computing $y^2$ for each element $y$ in the field---is still an efficient approach. Using that $(-y)^2 = y^2$ cuts our work about in half, leaving us to compute $0^2, 1^2, \ldots, 9^2$. One can check this using this table of quadratic residues for small primes.
A: Quadratic reciprocity has been mentioned. It is not really a great saving when $p =19$, but given a general prime $p$, to check which numbers are quadratic residues 
(mod $p$), it suffices to understand which primes $q < p$ are quadratic residues.
In fact, since $-1$ is a quadratic residue if $p \equiv 1$ (mod $4$) and a quadratic non-residue otherwise it suffices to check $q \leq \frac{p-1}{2}$.
To illustrate with $p = 19$. Since $19 \equiv 3$ (mod $8$), $2$ is a non-residue.
Since $19 \equiv 1$ (mod $3$) we see that $3$ is a non-residue (mod $19$). Since $19 \equiv 4$ (mod $5$), we see that $5$ is a residue (mod $19$). Since $19$ is a non-residue (mod $7$), we see that $7$ is a residue (mod $19$).
From this we can conclude (+ denotes square, - denotes non-square):
1 +, 2-,3-,4+,5+,6+,7+,8-,9+, 10-,11+,12-,13-,14-,15-,16+,17+,18-.
Another general procedure is that $n $ is a quadratic residue (mod $p$)  (for $1 \leq n <p$) if and only if $n^{\frac{p-1}{2}} \equiv 1$ (mod $p$), and this can be calculated relatively quickly.
Finally, squaring numbers consecutively (and taking residues (mod $p$) when $p = 19$) leads to $1,4,9,16,6,17,11,7,5$ being the only squares in $\mathbb{F}_{19}$, which is consistent with the results obtained by quadratic reciprocity.
