Show that $\overline{(2+\sqrt{3})}^{17} = \overline{2 - \sqrt{3}}$ in $\mathbb{Z}[\sqrt{3}]/(17)$ I am trying to solve the following problem:
Show that $\overline{(2+\sqrt{3})}^{17} = \overline{2 - \sqrt{3}}$ and $\overline{(2-\sqrt{3})}^{17} = \overline{2 + \sqrt{3}}$ in $\mathbb{Z}[\sqrt{3}]/(17)$, being $(17)$ the principal idela generated by $17$ in $\mathbb{Z}[\sqrt{3}]$ and $\overline{\alpha}$ the equivalence class of $\alpha$ in this quotient.
that is an intermediate step to calculating the remainder of $(2+\sqrt{3})^{17} + (2-\sqrt{3})^{17}$ in the division by 17.
My solution:
I am having some trouble on figuring out what is the set $\mathbb{Z}[\sqrt{3}]/(17)$. As I understand, given some $\alpha = a + b\sqrt{3} \in \mathbb{Z}[\sqrt{3}]$, then:
$$
\overline{\alpha} = \overline{a+b\sqrt{3}} = a + b\sqrt{3} + 17\mathbb{Z}[\sqrt{3}] = (a + 17\mathbb{Z}) + (b\sqrt{3} + 17\mathbb{Z}\sqrt{3}) = \overline{a} + \overline{b}\sqrt{3}
$$
where $\overline{a}$ is the equivalence class of $a$ in $\mathbb{Z}/17\mathbb{Z}$.
Question 1: is the above correct? If not, what is $\overline{\alpha} \in \mathbb{Z}[\sqrt{3}]/(17)$?
Then, assuming the above was right, i performed the following calculations:
$$
(\overline{2} + \overline{1}\sqrt{3})^2 = \overline{7} + \overline{2}\sqrt{3} 
$$
$$
\therefore (\overline{2} + \overline{1}\sqrt{3})^3 = \overline{3}(\overline{1} - \overline{2}\sqrt{3})
$$
But then
$$
(\overline{2} + \overline{1}\sqrt{3})^{17} = (\overline{2} + \overline{1}\sqrt{3})^2(\overline{1} - \overline{2}\sqrt{3})^5\overline{3}^5 = \overline{11} - \overline{7}\sqrt{3}
$$
In this last step I omitted some computations that I believe are not important to show here, since I only wanted to show my idea for solving the problem.
Question 2: Is the above idea correct? If not, how to solve the problem?
Any hints and comments will be the most appreciated.
 A: Without answering your actual questions, here is another approach to the problem.
We have $\mathbb Z[\sqrt 3]/(17)\simeq \mathbb Z[x]/(x^2-3,17)\simeq \mathbb Z/(17)[x]/(x^2-3)$.  We are looking to simplify $(2+x)^{17}$ in this ring.  But because $17$ is prime, all the binomial coefficients $\binom{17}{i}$ with $0<i<17$ will be multiples of $17$ (if you have not come across this fact, try to prove it), and so in our ring, $(2+x)^{17}=2^{17}+x^{17}$ by the binomial theorem.  By Fermat's little theorem, $2^{17}=2\pmod{17}$, and so $2^{17}+x^{17}=2+3^{8}x$, where we used $x^2=3$.  Our calculation is finished by computing $3^{8}\equiv 9^{4}\equiv (-4)^2\equiv -1 \pmod{17}$, which gets you the answer you want.
Alternatively, we can compute $3^8\pmod{17}$ by quadratic reciprocity by noting that $$3^{8}\equiv 3^{(17-1)/2}\equiv\left(\frac{3}{17} \right)\equiv (-1)^{(3-1)(17-1)/4}\left(\frac{17}{3} \right)\equiv\left(\frac{2}{3} \right)\equiv -1\pmod{17}$$
N.B. The fact that $(a+b)^p=a^p+b^p$ in a ring of characteristic $p$ is a standard trick, and it implies that the map $x\mapsto x^p$ is a ring automorphism called the Frobenius automorphism.  While it might look somewhat ad hoc here, it is a standard tool in the algebraist's toolbox. 
