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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 $7^2-3(16)=1$.

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Have you done Quadratic Reciprocity? –  André Nicolas Oct 10 '13 at 6:00
    
ah, yes. So, we have $(\frac{3}{p})=1$ and $(\frac{p}{3})=1$ for $p\equiv 1(mod 3)$ and $(\frac{p}{3})=-1$ for $p\equiv -1(mod 3)$. I'm not sure what to do next... –  Sarah Oct 10 '13 at 6:04
    
It is slightly more complicated. For $p\equiv 1\pmod{4}$, we have $(3/p)=(p/3)$. So if $p\equiv 1\pmod{4}$ and $p\equiv 1\pmod{3}$, that is, $p\equiv 1\pmod{12}$, we have $3$ is a QR. If $p\equiv 3\pmod{4}$, then $(3/p)=-(p/3)$. So $3$ is also a QR if $p\equiv 5\pmod{12}$ –  André Nicolas Oct 10 '13 at 6:10
    
Oh... I see, that makes sense. –  Sarah Oct 10 '13 at 6:15
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1 Answer

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:

$$T_{17}=\{(\pm1,0),(\pm2,\pm1),(\pm 5,\pm5),(\pm7,\pm 4),(\pm 8,\pm 2)\}$$

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