TITLE: Primes in a group: number theory
QUESTION [5 upvotes]: Characterize the primes for which 3 is not a square in $Z_p$. Compute $T_{17}$ where $T_p:=\{a+\sqrt{3}b:a^2-3b^2=1\} \subset Z_p[\sqrt{3}]$. Compute $T_{17}$. What are the orders of each of the elements in $T_{17}$?
I'm stuck on the first part.... I don't know how to really characterize these primes or what that means. 
For computing $T_{17}$, I am assuming I need to find the contained elements. Do I need to check for all possible combinations of a and b? or is there a shortcut of some sort?
For example, I can see that $a=7$, $b=4$ works with $7+4\sqrt{3}$ since $7^2-3(16)=1$.

REPLY [2 votes]: Case 1: $p\equiv 1\mod 4$
By quadratic reciprocity, we have $x^2\equiv 3\mod p$ is solvable if and only if $x^2\equiv p\mod 3$ is solvable.  The latter has a solution if and only if $p\equiv 1\mod 3$ (remembering that $p\ne 3$) so that in this case we have $p\equiv 1\mod 12$.
Case 2: $p \equiv 3\mod 4$
By quadratic reciprocity, we have $x^2\equiv 3\mod p$ is solvable if and only if $x^2\equiv p\mod 3$ is not solvable.  Again, the latter has a solution if and only if $p\equiv 1\mod 3$, so that in this case the original congruence has a solution if and only if $p\equiv 11\mod 12$.
The two cases show that the primes such that $3$ is not a square in $\mathbb{Z}_p$ are exactly those congruent to $5$ and $7$ modulo $12$.
To compute $T_{17}$, you could use symmetry to only check values $0,1,\ldots, 8$ because if $(a,b)$ is a solution to $a^2-3b^2=1$, then so is $(\pm a,\pm b)$.  In any case, we obtain the following $18$ elements, recorded as ordered pairs:
$$T_{17}=\{(\pm1,0),(\pm2,\pm1),(\pm 5,\pm5),(\pm7,\pm 4),(\pm 8,\pm 2)\}$$