Let $F$ be a non-archimedean local field of residual characteristic $p\neq 2$, and let $E=F[\sqrt{\epsilon}]$ be the quadractic unramified extension, here $\epsilon$ is a non-square element of $\mathcal{O}_{F}^{\times}$. Let $a\in\mathcal{O}_{E}^{\times}$, and $b\in\mathfrak{p}_{E}$ such that
-
$a\overline{a}=1-b\overline{b}$ (here for $a:=a_1+\sqrt{\epsilon}a_2,\;\overline{a}=a_1-\sqrt{\epsilon}a_2$),
-
$\overline{a}b+a\overline{b}=0$.
Then, is it possible to prove $a\in E^{1}(1+\mathfrak{p}_{E})$?
(First approach, given $b$, one can try to find the number of possible elements $a$ such that the above two equations hold. If we take $a=a_1+\sqrt{\epsilon}a_2$, and $b=b_1+\sqrt{\epsilon}b_2$, then Equations 1 and 2, give the relation $(a_1+b_1)^{2}-\epsilon(a_2+b_2)^{2}=1$….)
(Second approach, since $1-b\overline{b}\in 1+\mathfrak{p}_{F}^{2}$, there exists a $c\in\mathcal{O}_{F}^{\times}$ such that $c^2=1-b\overline{b}\in 1+\mathfrak{p}_{F}^{2}$, let $\lambda\in E^1$, then if we take $a=c\lambda$, then $a$ satisfies Equation 1. For it to satisfy Equation 2 as well we must have
$$\overline{c\lambda}b+c\lambda \overline{b}=0
\implies \overline{c}b+c\overline{b}=0\implies c\overline{b}\in\sqrt{\epsilon}\mathcal{O}_{F}.$$
Then we have $c=\smash{\overline{b}}^{-1}\sqrt{\epsilon}u$ for some $u\in\mathcal{O}_{F}$. Upon solving $c\overline{c}=1$, we get $-\epsilon^{-1}b\overline{b}=u^2$.
If $-\epsilon^{-1}b\overline{b}$ is a square in $F$, then we have two choices of $u$.)
Not sure about the next step
Best Answer
You only need (1). A Hensel's-lemma type approximation, using surjectivity of $\operatorname{tr}_{k_F/k_E}$ (where $k$ denotes residue fields), shows that $N_{E/F} : 1 + \mathfrak p_E \to 1 + \mathfrak p_F$ is surjective. Therefore, since $N_{E/F}(a) = 1 - N_{E/F}(b)$ belongs to $1 + \mathfrak p_F$, there is some $c \in 1 + \mathfrak p_E$ such that $N_{E/F}(c)$ equals $N_{E/F}(a)$, and then $a c^{-1}$ belongs to $E^1$.