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.

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. 
