Dihedral extension of 2-adic number field Sorry if the question is too long and maybe elementary.
I am reading a paper by Hirotada Naito on "Dihedral extensions of degree 8 over the rational p-adic fields". To generate dihedral extension $K(\sqrt{\epsilon},\sqrt{\epsilon^\sigma})$ in part 2-1, he said let $K=\mathbb{Q}_2(\sqrt{m})$ for $m=2,-2, 10$ or $-10$ 
$$K^*/K^{*2}\cong (\langle\sqrt{m}\rangle/\langle m\rangle)\times (\mathcal{O}^\times/\langle 1+m+2\sqrt{m}, 1+ \mathfrak{p}^5 \rangle)$$ 
for $\mathfrak{p}= \langle\sqrt{m}\rangle$. 
And after that, he said "it is sufficient to examine $\epsilon$ and $\epsilon\sqrt{m}$ where $\epsilon=a+b\sqrt{m}$ for $a=1,3,5,7$ and $b=0,1,2,3$."
My first question is "why it is enough to check these numbers?"
In the following, he said "we take $\epsilon$ (resp. $\epsilon\sqrt{m}$) such that $\epsilon$, $\epsilon^\sigma$, $\epsilon (1+m+2\sqrt{m})$ and $\epsilon^\sigma (1+m+2\sqrt{m})$ (resp. $\epsilon$, $-\epsilon^\sigma$, $\epsilon (1+m+2\sqrt{m})$ and $-\epsilon^\sigma (1+m+2\sqrt{m})$) are different modulo $\mathfrak{p}^5$ each other." Where $\sigma$ is the generator of Galois group of $K/\mathbb{Q}_2$.
And in the following, for $m=2$ he took $1+\sqrt{2}$, $3+\sqrt{2}$, $\sqrt{2}$ and $3\sqrt{2}$. I couldn't get to these numbers following his method explained above, can anyone show me the calculations? And can you show me any other possible numbers other than the given ones?
Here is the paper: Dihedral extensions of degree 8 over the rational p-adic fields.
Thank you. 

Answer to your first question: The numbers contain a set of representatives of $\mathfrak{o}^\times/(\mathfrak{o}^\times)^2$. (Naito said earlier that "we examine a representative system of $K_i^*/(K_i^*)^2$", and you copied his equation that shows representatives of this quotient group can be obtained from  representatives of $\mathfrak{o}^\times/(\mathfrak{o}^\times)^2$ and their multiples by $\sqrt{m}$.)
It is easy to see that these numbers contain a set of representatives if you note that $\mathfrak{p}^5 = 4\mathfrak{p}$. (This is obvious for $m = \pm 2$. For $m = \pm 10$, note that $5$ is a unit in $\mathfrak{o}$.)
For any $x \in \mathfrak{o}^\times$, write $x^{-1} = a' + b' \sqrt{m}$, and choose $a \equiv a' \pmod{8}$ and $b \equiv b' \pmod{4}$. Then it is easy to see that $x(a + b \sqrt{m}) \in 1 + 8 \mathfrak{o} + 4\mathfrak{p} \subseteq \mathfrak{o}^{\times 2}$ (since $1 + 8 \mathfrak{o} \subseteq (\mathfrak{o}^\times)^2$).

Partial answer to your second question:  We are supposed to check all 16 possible combinations of $a$ and $b$ (and also consider multiplying each possibility by $\sqrt{m}$). I will only do two of them (and I won't multiply by $\sqrt{m}$).
For $a = b = 1$ (with $m = 2$), we have:


*

*$\epsilon = 1 + \sqrt{2}$,

*$\epsilon^\sigma = 1 - \sqrt{2}$,

*$\epsilon(1+m+2 \sqrt{2}) = (1 + \sqrt{2})(3+2 \sqrt{2}) = 7 + 5 \sqrt{2}$,

*$\epsilon^\sigma(1+m+2 \sqrt{2}) = (1 - \sqrt{2})(3+2 \sqrt{2}) = -1 - \sqrt{2}$.


It is easy to check that none of these are congruent modulo $4 \sqrt{2}$:
$$x + y \sqrt{2} \equiv x' + y' \sqrt{2} \pmod{4 \sqrt{2}} \iff \text{$x \equiv x' \pmod{8}$ and $y \equiv y' \pmod{4}$} .$$
 So this gives an example, which means that $1 + 2 \sqrt{2}$ should be on Naito's list. It is indeed there.
For $a = 1$ and $b = 2$ (with $m = 2$), we have:


*

*$\epsilon = 1 + 2\sqrt{2}$,

*$\epsilon^\sigma = 1 - 2\sqrt{2}$,

*$\epsilon(1+m+2 \sqrt{2}) = (1 + 2\sqrt{2})(3+2 \sqrt{2}) = 11 + 8 \sqrt{2}$,

*$\epsilon^\sigma(1+m+2 \sqrt{2}) = (1 - 2\sqrt{2})(3+2 \sqrt{2}) = -1 - 2\sqrt{2}$.


Since 
    $$\epsilon - \epsilon^\sigma = (1 + 2\sqrt{2}) - (1 - 2 \sqrt{2}) = 4 \sqrt{2} \equiv 0 \pmod{4 \sqrt{2}} ,$$
this does not give an example. So $1 + 2 \sqrt{2}$ should not be on Naito's list. It is indeed not there.
By checking each of the 32 possibilities for each of the four possible values of $m$, it should be easy to determine whether his list is correct.
