# On the sum of two integral squares in certain quadratic fields

Dasheng Wei  Academy of Mathematics and System Science, CAS, Beijing 100190,
P.R.China<EMAIL_ADDRESS>

## Abstract

We propose a method for determining which integers can be written as a sum of
two integral squares for certain quadratic fields.

MSC classification : 11E12;11D09

Keywords : integral points, ring class field, reciprocity law.

## Introduction

It is significantly more difficult to study sums of two integral squares over
algebraic number fields than that over ${\mathbb{Z}}$. From nowadays point of
view, Fermat-Gauss’ theorem about sums of two squares over ${\mathbb{Z}}$ is a
purely local problem. However such question is a global problem over algebraic
number fields, even quadratic fields since the class number of number fields
is involved. There are only a few results about the question for general
algebraic number fields $F$. Niven studied the problem for
$F={\mathbb{Q}}({\sqrt{-1}})$ in [Niv]. This case is very special since the
binary quadratic form $x^{2}+y^{2}$ is hyperbolic over
${\mathbb{Q}}(\sqrt{-1})$. Nagell further studied the problem for
$F={\mathbb{Q}}(\sqrt{d})$ when $d$ is one of the following twenty integers:

$\pm 2,\pm 3,\pm 5,\pm 7,\pm 11,\pm 13,\pm 19,\pm 43,\pm 67,\pm 163$

in [Nag1] and [Nag2]. His method follows Gauss’ original idea and essentially
depends on the fact that the class number of
${\mathbb{Q}}(\sqrt{d},\sqrt{-d})$ is 1 when $d$ is one of the above integers.
Obviously this method can not apply for general algebraic number fields.

Recently Harari (in [Ha08]) proved that the Brauer-Manin obstruction to the
existence of an integral point is the only obstruction for an integral model
of a principal homogenous space of tori. However, the Brauer group is
infinite. One cannot use these results to determine the existence of integral
points on a specific example. Fei Xu and the author gave another proof of the
result in [WX] and [WX2]. In this paper we apply the method in [WX] for the
sum of two squares over quadratic fields.

It should be pointed out that the method in [WX] only produces the idelic
class groups of $F(\sqrt{-1})$ for solving the problem of sum of two squares.
And these idelic class groups are not unique. In order to get the explicit
conditions for the sum of two squares, one needs further to construct the
explicit abelian extensions of $F(\sqrt{-1})$ corresponding to the idelic
class groups by the class field theory. Such explicit construction is a wide
open problem (Hilbert’s 12-th problem) in general but ad hoc methods.

Notation and terminology are standard if not explained. Let $F$ be a number
field, $\mathfrak{o}_{F}$ the ring of integers of $F$, $\Omega_{F}$ the set of
all primes in $F$ and $\infty$ the set of all infinite primes in $F$. For
simplicity, we write $\mathfrak{p}<\infty$ for
$\mathfrak{p}\in\Omega_{F}\setminus\infty$. Let $F_{\mathfrak{p}}$ be the
completion of $F$ at $\mathfrak{p}$ and $\mathfrak{o}_{F_{\mathfrak{p}}}$ be
the local completion of $\mathfrak{o}_{F}$ at $\mathfrak{p}$ for each
$\mathfrak{p}\in\Omega_{F}$. Write
$\mathfrak{o}_{F_{\mathfrak{p}}}=F_{\mathfrak{p}}$ for
$\mathfrak{p}\in\infty$. We also denote the adele ring (resp. the idele ring)
of $F$ by $\mathbb{A}_{F}$ (resp. $\mathbb{I}_{F}$).

Suppose that $-1$ is not a square in $F$. Let $E=F(\sqrt{-1})$. Let
$\mathbb{X}_{\alpha}$ denote the affine scheme over $\mathfrak{o}_{F}$ defined
by the equation $x^{2}+y^{2}=\alpha$ for a non-zero integer
$\alpha\in\mathfrak{o}_{F}$. The equation $x^{2}+y^{2}=\alpha$ is solvable
over $\mathfrak{o}_{F}$ if and only if
$\mathbb{X}_{\alpha}(\mathfrak{o}_{F})\neq\emptyset$. Let
$X_{\alpha}=\mathbb{X}_{\alpha}\times_{\mathfrak{o}_{F}}{F}$. Obviously
$f=x+y\sqrt{-1}$ is an invertible function on $X_{\alpha}\otimes_{F}E$. And
$f$ induces a natural map

$f_{E}:\ \ \ X_{\alpha}(\mathbb{A}_{F})\rightarrow\mathbb{I}_{E}.$

The restriction to $X_{\alpha}(F_{\mathfrak{p}})$ of $f_{E}$ can be defined by

$f_{E}[(x_{\mathfrak{p}},y_{\mathfrak{p}})]=\begin{cases}(x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},x_{\mathfrak{p}}-y_{\mathfrak{p}}\sqrt{-1})\
\ \ &\text{if $\mathfrak{p}$ splits in $E/F$,}\\\
x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1}\ \ \ &\text{otherwise.}\end{cases}$

######  Definition 0.1.

Let $K_{1},\cdots,K_{n}$ be finite abelian extensions over $E$. Let

$\psi_{K_{i}/E}:\mathbb{I}_{E}\rightarrow Gal(K_{i}/E)\text{ for }1\leq i\leq
n$

be the Artin map. We say that $\alpha$ satisfies the Artin condition of
$K_{1},\cdots,K_{n}$ if there is

$\prod_{\mathfrak{p}\leq\infty}(x_{\mathfrak{p}},y_{\mathfrak{p}})\in\prod_{\mathfrak{p}\leq\infty}\mathbb{X}_{\alpha}(\mathfrak{o}_{F_{\mathfrak{p}}})$

such that

$\psi_{K_{i}/E}(f_{E}[\prod_{\mathfrak{p}\leq\infty}(x_{\mathfrak{p}},y_{\mathfrak{p}})])=1_{i}\text{
for }i=1,\cdots,n$

where $1_{i}$ is the identity element of $Gal(K_{i}/E)$.

By the class field theory, it is a necessary condition for
$\mathbb{X}_{\alpha}(\mathfrak{o}_{F})\neq\emptyset$ that $\alpha$ satisfies
the Artin condition of $K_{1},\cdots,K_{n}$. There exists a (not unique)
finite abelian extension $K/E$ independent on $\alpha$, such that the Artin
condition of $K$ is also sufficient for
$\mathbb{X}_{\alpha}(\mathfrak{o}_{F})\neq\emptyset$ (Corollary 2.18 in [WX]).
Let $d\geq 2$ be a positive square-free integer and
$F={\mathbb{Q}}(\sqrt{-d})$. Then the field $K$ closely depends on the (local
and global) solvability of the following equation

$x^{2}+y^{2}=-1.$ (0.1)

Let $L=\mathfrak{o}_{F}+\mathfrak{o}_{F}\sqrt{-1}$ and $H_{L}$ be the ring
class field corresponding to the order $L$. For example, the Artin condition
of $H_{L}$ is sufficient if $F={\mathbb{Q}}(\sqrt{p})\text{ or
}{\mathbb{Q}}(\sqrt{-p})$ with $p$ prime (Theorem 0.3 in [Wei]). And the
following result can be found in [Wei] (Proposition 1.1).

######  Proposition 0.2.

Suppose one of the following conditions holds:

(1) The equation (0.1) has an integral solution in $\mathfrak{o}_{F}$.

(2) The equation (0.1) has no local integral solutions at a place of $F$.

Then the equation $x^{2}+y^{2}=\alpha$ is solvable over $\mathfrak{o}_{F}$ if
and only if $\alpha$ satisfies the Artin condition of $H_{L}$.

Therefore we only need to consider the case that the two conditions in
Proposition 0.2 don’t hold. For this case the Arin condition of $H_{L}$ must
not be sufficient. In general the question will be very complicated. We can
not expect to construct a field $K$ generally, such that the Artin condition
of $K$ is sufficient.

The following result is well-known (Satz 2, [Pe]): the equation (0.1) is
solvable over $\mathfrak{o}_{F}$ if and only if the equation

$x^{2}-dy^{2}=-\gamma(d)$ (0.2)

is solvable over ${\mathbb{Z}}$, where

$\gamma(d)=\begin{cases}1&\text{if }d\not\equiv-1\mod 4\\\ 2&\text{if
}d\equiv-1\mod 4.\end{cases}$

In this paper, we will consider the case that the equation (0.1) is solvable
over $\mathfrak{o}_{F_{\mathfrak{p}}}$ for any $\mathfrak{p}\in\Omega_{F}$ and
the equation (0.2) is not solvable over ${\mathbb{Z}}_{p}$ for some prime $p$.
It is easy to see that the two conditions in Proposition 0.2 don’t hold for
this case. Let

$\displaystyle C=\\{(d,p)|d\text{ is a square-free positive integer and }p\mid
d\text{ with }p\text{ prime}\\}$ $\displaystyle D_{1}=\\{(d,p)\in
C|d\not\equiv-1\mod 8,p\equiv-1\mod 8\\}$ $\displaystyle D_{2}=\\{(d,p)\in
C|d\equiv 1,2\mod 4,p\equiv 3\mod 8\\}$ $\displaystyle D_{3}=\\{(d,p)\in
C|d\equiv 3\mod 8,p\equiv 5\mod 8\\}$

Denote

$D=D_{1}\cup D_{2}\cup D_{3}.$

We say that $d\in D\ (\text{or }D_{i})$ if there is a prime $p$ such that
$(d,p)\in D\ (\text{or }D_{i})$. It is easy to verify that the equation (0.1)
is solvable over $\mathfrak{o}_{F_{\mathfrak{p}}}$ for any
$\mathfrak{p}\in\Omega_{F}$ and the equation (0.2) is not solvable over
${\mathbb{Z}}_{p}$ for some prime $p$ if and only if $d\in D$. In this paper,
we mainly prove the following result.

######  Theorem 0.3.

Let $(d,p)\in D$ and $F={\mathbb{Q}}(\sqrt{-d})$. Then the diophantine
equation $x^{2}+y^{2}=\alpha$ is solvable over $\mathfrak{o}_{F}$ if and only
if $\alpha$ satisfies the Artin condition of $\Theta$ and $H_{L}$,where
$\Theta=E(\sqrt[4]{p})$.

Sums of two integral squares over real quadratic fields is more complicated
than that over imaginary quadratic fields, since there are more units in real
quadratic fields. In this paper, we also obtain a result about real quadratic
fields.

######  Theorem 0.4.

Let $p$ be a prime and $F={\mathbb{Q}}(\sqrt{2p})$. If $p\equiv 3\mod 8$, then
the diophantine equation $x^{2}+y^{2}=\alpha$ is solvable over
$\mathfrak{o}_{F}$ if and only if $\alpha$ satisfies the Artin condition of
$\Theta$ and $H_{L}$, where $\Theta=E(\sqrt[4]{2})$.

As application, we explicitly determine which integers can be written as a sum
of two integral squares for the two quadratic fields ${\mathbb{Q}}(\sqrt{\pm
6})$, and ${\mathbb{Q}}(\sqrt{\pm 6})$ are the fields with the smallest $|d|$
that the Artin condition of $H_{L}$ is not sufficient for
${\mathbb{Q}}(\sqrt{d})$.

##  1\. The sum of two squares in imaginary quadratic fields

Let $F$ be an algebraic number field and $-1$ is not a square in $F$. Let
$E=F(\sqrt{-1})$ and let $T$ be the torus $R^{1}_{E/F}(\mathbb{G}_{m})={\rm
Ker}[R_{E/F}(\mathbb{G}_{m,E})\rightarrow\mathbb{G}_{m,F}]$, here $R$ denotes
the Weil’s restriction (see [Milne98]). Denote $\lambda$ to be the embedding
from $T$ to $R_{E/F}(\mathbb{G}_{m,E})$. Obviously $\lambda$ induces a natural
group homomorphism

$\lambda_{E}:\ \ \ T(\mathbb{A}_{F})\rightarrow\mathbb{I}_{E}.$

Let $\mathbb{X}_{\alpha}$ denote the affine scheme over $\mathfrak{o}_{F}$
defined by $x^{2}+y^{2}=\alpha$ for a non-zero integer
$\alpha\in\mathfrak{o}_{F}$. Let $\mathbb{T}$ be the group scheme over
$\mathfrak{o}_{F}$ defined by $x^{2}+y^{2}=1$ and let
$T=\mathbb{T}\times_{\mathfrak{o}_{F}}{F}$. The generic fiber of
$\mathbb{X}_{\alpha}$ is a principal homogenous space of the torus $T$. Since
$\mathbb{T}$ is separated over $\mathfrak{o}_{F}$, we can view
$\mathbb{T}(\mathfrak{o}_{F_{\mathfrak{p}}})$ as a subgroup of
$T(F_{\mathfrak{p}})$. The following result can be founded in [WX] (Corollary
2.20).

######  Proposition 1.1.

Let $K_{1}/E$ and $K_{2}/E$ be finite abelian extensions such that the group
homomorphism $\widetilde{\lambda}_{E}$ induced by $\lambda_{E}$

$\widetilde{\lambda}_{E}:\nicefrac{{T(\mathbb{A}_{F})}}{{T(F)\prod_{\mathfrak{p}\leq\infty}\mathbb{T}(\mathfrak{o}_{F_{\mathfrak{p}}})}}\longrightarrow\nicefrac{{\mathbb{I}_{E}}}{{E^{*}N_{K_{1}/E}(\mathbb{I}_{K_{1}})}}\times\nicefrac{{\mathbb{I}_{E}}}{{E^{*}N_{K_{2}/E}(\mathbb{I}_{K_{2}})}}$

is well-defined and injective, where well-defined means

$\lambda_{E}(T(F)\prod_{\mathfrak{p}\leq\infty}\mathbb{T}(\mathfrak{o}_{F_{\mathfrak{p}}}))\subset(E^{*}N_{K_{1}/E}(\mathbb{I}_{K_{1}}))\cap(E^{*}N_{K_{2}/E}(\mathbb{I}_{K_{2}})).$

Then $\mathbb{X}_{\alpha}(\mathfrak{o}_{F})\neq\emptyset$ if and only if
$\alpha$ satisfies the Artin condition of $K_{1}$ and $K_{2}$.

Let $d\geq 2$ be a square-free positive integer. Let
$F={\mathbb{Q}}(\sqrt{-d})$, $\mathfrak{o}_{F}$ be the ring of integers of $F$
and $E=F(\sqrt{-1})$. In the rest of this section $F$ is always an imaginary
quadratic field. One takes the order
$L=\mathfrak{o}_{F}+\mathfrak{o}_{F}\sqrt{-1}$ inside $E$. Let $H_{L}$ be the
ring class field corresponding to the order
$L=\mathfrak{o}_{F}+\mathfrak{o}_{F}\sqrt{-1}$. Recall some notations:

$\displaystyle C=\\{(d,p)|d\text{ is a square-free positive integer and }p\mid
d\text{ with }p\text{ prime}\\}$ $\displaystyle D_{1}=\\{(d,p)\in
C|d\not\equiv-1\mod 8,p\equiv-1\mod 8\\}$ $\displaystyle D_{2}=\\{(d,p)\in
C|d\equiv 1,2\mod 4,p\equiv 3\mod 8\\}$ $\displaystyle D_{3}=\\{(d,p)\in
C|d\equiv 3\mod 8,p\equiv 5\mod 8\\}.$

Note that $p\equiv 3\mod 4$ if $(d,p)\in D_{1}\cup D_{2}$.

######  Proposition 1.2.

Suppose $(d,p)\in D_{1}\cup D_{2}$ and $F={\mathbb{Q}}(\sqrt{-d})$. Let
$(x_{\mathfrak{p}},y_{\mathfrak{p}})\in\mathfrak{o}_{F_{\mathfrak{p}}}\times\mathfrak{o}_{F_{\mathfrak{p}}}$,
then the $4$-th Hilbert symbol

$\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},p}{v}\right)_{4}=\begin{cases}1&\text{if
}x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=1\\\ -1&\text{if
}x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=-1,\end{cases}$

where $v$ and $\mathfrak{p}$ are respectively the unique place of $E$ and $F$
above $p$.

###### Proof.

The Hilbert symbol

$\displaystyle\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},p}{v}\right)_{4}$
$\displaystyle=\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},-d}{v}\right)_{4}\cdot\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},-d/p}{v}\right)^{-1}_{4}$
$\displaystyle=\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},-d}{v}\right)_{4}\cdot
1=\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},\sqrt{-d}}{v}\right)$
$\displaystyle=\left(\frac{\pm 1,p}{p}\right)=\pm 1,$

where the second equation holds since $E(\sqrt[4]{-d/p})/E$ is unramified at
$v$. ∎

######  Proposition 1.3.

Suppose $(d,p)\in D_{1}\cup D_{2}$ and $F={\mathbb{Q}}(\sqrt{-d})$. If
$x_{\mathfrak{p}}$ and $y_{\mathfrak{p}}$ in $\mathfrak{o}_{F_{\mathfrak{p}}}$
satisfy $x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=\pm 1$, then

$\prod_{v|2}\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},p}{v}\right)_{4}=1$

where $v\in\Omega_{E}$ and $\mathfrak{p}$ is the unique place of $F$ above
$2$.

The proposition follows from the series of lemmas.

######  Lemma 1.4.

Let $d\not\equiv-1\mod 8$ and $l\equiv-1\mod 8$. Let
$F={\mathbb{Q}}(\sqrt{-d})$. If $x_{\mathfrak{p}}$ and $y_{\mathfrak{p}}$ in
$F_{\mathfrak{p}}$ satisfy $x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=\pm 1$,
then

$\prod_{v|2}\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},l}{v}\right)_{4}=1$

where $v\in\Omega_{E}$ and $\mathfrak{p}$ is the unique place of $F$ above
$2$.

###### Proof.

The extension $E(\sqrt{l})/E$ is split over $v$. And

$\sqrt{l}=s\sqrt{-1}=(2\sqrt{-1})\cdot\frac{s}{2}=(1+\sqrt{-1})^{2}\frac{2s}{2^{2}},$

where $s^{2}=-l$ and $s\in{\mathbb{Z}}^{\times}_{2}$. For any
$x_{\mathfrak{p}}$ and $y_{\mathfrak{p}}$ in $F_{\mathfrak{p}}$ satisfy
$x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=\pm 1$, we have

$\displaystyle\prod_{v|2}\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},l}{v}\right)_{4}$
$\displaystyle=\prod_{v|2}\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},\sqrt{l}}{v}\right)=\prod_{v|2}\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},2s}{v}\right)$
$\displaystyle=\left(\frac{\pm 1,2s}{\mathfrak{p}}\right)=1.$

∎

Proposition 1.3 for the case $(d,p)\in D_{1}$ follows from the above lemma.
The following two lemmas deal with the case $(d,p)\in D_{2}$.

######  Lemma 1.5.

Let $d\equiv 1\mod 4$ and $l\equiv 3\mod 8$. Let $F={\mathbb{Q}}(\sqrt{-d})$.
If $x_{\mathfrak{p}}$ and $y_{\mathfrak{p}}$ in
$\mathfrak{o}_{F_{\mathfrak{p}}}$ satisfy
$x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=\pm 1$, then

$\prod_{v|2}\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},l}{v}\right)_{4}=1$

where $v\in\Omega_{E}$ and $\mathfrak{p}$ is the unique place of $F$ above
$2$.

###### Proof.

(1) If $d\equiv 5\mod 8$, then
$F={\mathbb{Q}}(\sqrt{-d})={\mathbb{Q}}(\sqrt{l})$. And $v$ is the unique
place of $E$ above $2$. For any $x_{\mathfrak{p}}$ and $y_{\mathfrak{p}}$ in
$F_{\mathfrak{p}}$ satisfy $x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=\pm 1$,
we have

$\displaystyle\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},l}{v}\right)_{4}$
$\displaystyle=\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},\sqrt{l}}{v}\right)$
$\displaystyle=\left(\frac{\pm
1,\sqrt{l}}{\mathfrak{p}}\right)=\left(\frac{\pm 1,-l}{2}\right)$
$\displaystyle=1.$

(2) If $d\equiv 1\mod 8$, then $E/F$ is split over $\mathfrak{p}$. For any
$x_{\mathfrak{p}}$ and $y_{\mathfrak{p}}$ in $\mathfrak{o}_{F_{\mathfrak{p}}}$
satisfy $x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=\pm 1$, we have

$\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},l}{v}\right)=1$

since $E(\sqrt{l})/E$ is unramified at any place $v$ of $E$ above $2$. So we
have

$\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},l}{v}\right)_{4}=\pm
1.$

Therefore

$\displaystyle\prod_{v|2}\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},l}{v}\right)_{4}$
$\displaystyle=\left(\frac{(x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1})(x_{\mathfrak{p}}-y_{\mathfrak{p}}\sqrt{-1}),l}{\mathfrak{p}}\right)_{4}$
$\displaystyle=\begin{cases}1&\text{ if
}x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=1\\\
\left(\frac{-1,l}{\mathfrak{p}}\right)_{4}=\left(\frac{\sqrt{-1},l}{\mathfrak{p}}\right)=1&\text{
if }x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=-1.\end{cases}$

∎

######  Lemma 1.6.

Let $d\equiv 2\mod 4$ and $l\equiv\pm 3\mod 8$. Let
$F={\mathbb{Q}}(\sqrt{-d})$. If $x_{\mathfrak{p}}$ and $y_{\mathfrak{p}}$ in
$\mathfrak{o}_{F_{\mathfrak{p}}}$ satisfy
$x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=\pm 1$, then

$\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},l}{v}\right)_{4}=1$

where $v$ and $\mathfrak{p}$ are respectively the unique place of $E$ and $F$
above $2$.

###### Proof.

(1) First we will prove that
$\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},l}{v}\right)_{4}=1$
for any $x_{\mathfrak{p}}$ and $y_{\mathfrak{p}}$ in
$\mathfrak{o}_{F_{\mathfrak{p}}}$ that satisfy
$x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=1$.

By Hilbert 90, there exists $\beta\in E_{v}^{*}$ such that
$x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1}=\sigma(\beta)/\beta$, where
$\sigma$ is the non-trivial element of $Gal(E_{v}/F_{\mathfrak{p}})$. Let
$\beta=a+b\sqrt{-1}$ and we can choose $\beta$ such that
$a,b\in\mathfrak{o}_{F_{\mathfrak{p}}}$ and $a$ or $b$ is a unit. Denote
$\kappa=a^{2}+b^{2}$.

Assume that $ord_{\mathfrak{p}}(\kappa)$ is odd. The equation
$x^{2}+y^{2}=\kappa$ is not solvable over $\mathfrak{o}_{F_{\mathfrak{p}}}$ if
$ord_{\mathfrak{p}}(\kappa)=1$ (by Theorem 1 in [Ko]). Therefore
$ord_{\mathfrak{p}}(\kappa)\geq 3$. We have
$a,b\in\mathfrak{o}_{F_{p}}^{\times}$ and

$\beta/\sigma(\beta)=(a^{2}-b^{2}-2ab\sqrt{-1})\kappa^{-1}=1-2\kappa^{-1}b^{2}-2\kappa^{-1}ab\sqrt{-1}.$

Since $ord_{\mathfrak{p}}(\kappa)\geq 3$ and
$a,b\in\mathfrak{o}_{F_{\mathfrak{p}}}^{\times}$, one has

$2\kappa^{-1}b^{2},2\kappa^{-1}ab\not\in\mathfrak{o}_{F_{\mathfrak{p}}}.$

So
$\beta/\sigma(\beta)\not\in\mathfrak{o}_{F_{\mathfrak{p}}}+\mathfrak{o}_{F_{\mathfrak{p}}}\sqrt{-1}$.
A contradiction is derived, so one obtains $ord_{\mathfrak{p}}(\kappa)$ is
even.

Let $ord_{\mathfrak{p}}(\kappa)=2n$. Since $E/F$ is totally ramified and of
degree $2$, one can write $\beta=\pi_{F}^{n}\mu$, where $\pi_{F}$ is a
uniformizer of $F$ and $\mu\in\mathfrak{o}_{E_{v}}^{\times}$. Then
$\sigma(\beta)/\beta=\sigma(\mu)/\mu$. Let $\mathfrak{q}$ be the unique place
of ${\mathbb{Q}}(\sqrt{-1})$ above $2$. We have

$\displaystyle\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},l}{v}\right)_{4}=$
$\displaystyle\left(\frac{\sigma(\mu)/\mu,l}{v}\right)_{4}=\left(\frac{N_{E_{v}/F_{\mathfrak{p}}}(\mu),l}{v}\right)_{4}\cdot\left(\frac{\mu,l}{v}\right)^{-1}$
$\displaystyle=\left(\frac{N_{E_{v}/{\mathbb{Q}}_{2}}(\mu),l}{\mathfrak{q}}\right)_{4}\cdot\left(\frac{N_{E_{v}/{\mathbb{Q}}_{2}}(\mu),l}{2}\right)^{-1}.$

Since $\mu\in\mathfrak{o}_{E_{v}}^{\times}$ and
$E_{v}={\mathbb{Q}}_{2}(\sqrt{-1},\sqrt{-d})$ with $d\equiv 2\mod 4$, one has

$N_{E_{v}/{\mathbb{Q}}_{2}}(\mu)=m^{2}\text{ for some
}m\in{\mathbb{Z}}_{2}^{\times}.$

So

$\displaystyle\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},l}{v}\right)_{4}=\left(\frac{m^{2},l}{\mathfrak{q}}\right)_{4}\cdot
1=\left(\frac{m,l}{\mathfrak{q}}\right)=1.$

(2) Let $d=2d_{0}$. Then $d_{0}$ is odd. By the argument in (1), we only need
to show that there exists a
$(x_{\mathfrak{p}},y_{\mathfrak{p}})\in\mathfrak{o}_{F_{\mathfrak{p}}}\times\mathfrak{o}_{F_{\mathfrak{p}}}$
satisfying $x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=-1$, such that

$\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},l}{v}\right)_{4}=1.$

(i) If $d_{0}\equiv 1\mod 4$, then $x^{2}-2d_{0}y^{2}=-1$ is solvable over
${\mathbb{Z}}_{2}$. Choose one solution
$(x_{0},y_{0})\in{\mathbb{Z}}_{2}\times{\mathbb{Z}}_{2}$, then we have
$x_{0}^{2}+(y_{0}\sqrt{-2d_{0}})^{2}=-1$. Let

$x_{\mathfrak{p}}=x_{0}\text{ and }y_{\mathfrak{p}}=y_{0}\sqrt{-2d_{0}}.$

One obtains

$\displaystyle\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},l}{v}\right)_{4}$
$\displaystyle=\left(\frac{x_{0}-y_{0}\sqrt{2d_{0}},l}{v}\right)_{4}=\left(\frac{-1,l}{\mathfrak{q}}\right)_{4}=\left(\frac{\sqrt{-1},l}{\mathfrak{q}}\right)=1.$

Let $s\in{\mathbb{Z}}_{2}^{\times}$ such that $s^{2}=l/3\text{ or }-l/3$. For
any $\delta\in E_{v}^{*}$ satisfies $N_{E_{v}/F_{\mathfrak{p}}}(\delta)=\pm
1$, we have

$\left(\frac{\delta,l}{v}\right)_{4}=\left(\frac{\delta,3}{v}\right)_{4}\cdot\left(\frac{\delta,s}{v}\right)\cdot\left(\frac{\delta,\pm
1}{v}\right)_{4}=\left(\frac{\delta,3}{v}\right)_{4}\cdot\left(\frac{\pm
1,s}{\mathfrak{p}}\right)\cdot 1=\left(\frac{\delta,3}{v}\right)_{4}.$

(ii) If $d_{0}\equiv 3\mod 8$, then
$F_{\mathfrak{p}}={\mathbb{Q}}_{2}(\sqrt{-6})$. Let

$x_{\mathfrak{p}}=(1+2\sqrt{-6})/5\text{ and
}y_{\mathfrak{p}}=(2-\sqrt{-6})/5.$

Then $x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=-1$. One has

$\displaystyle\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},l}{v}\right)_{4}$
$\displaystyle=\left(\frac{5^{-1}(1+2\sqrt{-1})+5^{-1}(2-\sqrt{-1})\sqrt{-6},3}{v}\right)_{4}$
$\displaystyle=\left(\frac{5^{-1}(3-4\sqrt{-1}),3}{\mathfrak{q}}\right)_{4}$
$\displaystyle=\left(\frac{3-4\sqrt{-1},3}{\mathfrak{q}}\right)_{4}\cdot\left(\frac{5,3}{\mathfrak{q}}\right)_{4}^{-1}.$

It’s easy to see

$\left(\frac{5,3}{\mathfrak{q}}\right)_{4}=\left(\frac{-3,3}{\mathfrak{q}}\right)_{4}=1.$

By the class field theory, we have

$\displaystyle\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},l}{v}\right)_{4}$
$\displaystyle=\prod_{w|3}\left(\frac{3-4\sqrt{-1},3}{w}\right)_{4}\prod_{w|5}\left(\frac{3-4\sqrt{-1},3}{w}\right)_{4}$
$\displaystyle=(-\sqrt{-1})^{(3^{2}-1)/4}\cdot\left(\frac{5^{2},3}{5}\right)_{4}$
$\displaystyle=(-1)\cdot(-1)=1,$

where $w$ is in the set of places of ${\mathbb{Q}}(\sqrt{-1})$.

(iii) If $d_{0}\equiv 7\mod 8$, then
$F_{\mathfrak{p}}={\mathbb{Q}}_{2}(\sqrt{-14})$. Let

$x_{\mathfrak{p}}=(3-2\sqrt{-14})/13\text{ and
}y_{\mathfrak{p}}=(2+3\sqrt{-14})/13.$

Then $x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=-1$. One has

$\displaystyle\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},l}{v}\right)_{4}$
$\displaystyle=\left(\frac{13^{-1}(3+2\sqrt{-1})+13^{-1}(-2+3\sqrt{-1})\sqrt{-14},3}{v}\right)_{4}$
$\displaystyle=\left(\frac{13^{-1}(-5-12\sqrt{-1}),3}{\mathfrak{q}}\right)_{4}$
$\displaystyle=\left(\frac{-5-12\sqrt{-1},3}{\mathfrak{q}}\right)_{4}\cdot\left(\frac{13,3}{\mathfrak{q}}\right)_{4}^{-1}.$

It’s easy to see

$\left(\frac{13,3}{\mathfrak{q}}\right)_{4}=\left(\frac{-3,3}{\mathfrak{q}}\right)_{4}=1.$

By the class field theory, we have

$\displaystyle\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},l}{v}\right)_{4}$
$\displaystyle=\prod_{w|3}\left(\frac{-5-12\sqrt{-1},3}{w}\right)_{4}\prod_{w|13}\left(\frac{-5-12\sqrt{-1},3}{w}\right)_{4}$
$\displaystyle=1\cdot\left(\frac{13^{2},3}{13}\right)_{4}=1\cdot 1=1,$

where $w$ is in the set of places of ${\mathbb{Q}}(\sqrt{-1})$. ∎

######  Remark.

In the above three lemmas we don’t assume that $d>0$, $l$ prime and $l\mid d$.

Recall

$D_{3}=\\{(d,p)\in C|d\equiv 3\mod 8,p\equiv 5\mod 8\\}.$

In the following, we give some properties about the case $(d,p)\in D_{3}$.

######  Proposition 1.7.

Let $(d,p)\in D_{3}$ and $F={\mathbb{Q}}(\sqrt{-d})$. If $x_{\mathfrak{p}}$
and $y_{\mathfrak{p}}$ in $\mathfrak{o}_{F_{\mathfrak{p}}}$ satisfy
$x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=\pm 1$, then

$\prod_{v|p}\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},p}{v}\right)_{4}=1$

where $v\in\Omega_{E}$ and $\mathfrak{p}$ is the unique place of $F$ above
$p$.

###### Proof.

The Hilbert symbol

$\displaystyle\prod_{v|p}\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},p}{v}\right)_{4}$
$\displaystyle=\prod_{v|p}\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},-d}{v}\right)_{4}\cdot\prod_{v|p}\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},-d/p}{v}\right)^{-1}_{4}$
$\displaystyle=\prod_{v|p}\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},-d}{v}\right)_{4}\cdot
1=\prod_{v|p}\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},\sqrt{-d}}{v}\right)$
$\displaystyle=\left(\frac{\pm 1,p}{p}\right)=1,$

where the second equation holds since $E(\sqrt[4]{-d/p})/E$ is unramified over
$v$. ∎

######  Proposition 1.8.

Let $(d,p)\in D_{3}$ and $F={\mathbb{Q}}(\sqrt{-d})$. Let
$(x_{\mathfrak{p}},y_{\mathfrak{p}})\in\mathfrak{o}_{F_{\mathfrak{p}}}\times\mathfrak{o}_{F_{\mathfrak{p}}}$,
then

$\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},p}{v}\right)_{4}=\begin{cases}1&\text{if
}x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=1\\\ -1&\text{if
}x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=-1\end{cases}$

where $v$ and $\mathfrak{p}$ are respectively the unique place of $E$ and $F$
above $2$.

###### Proof.

The Hilbert symbol

$\displaystyle\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},p}{v}\right)_{4}$
$\displaystyle=\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},\sqrt{p}}{v}\right)=\left(\frac{\pm
1,\sqrt{p}}{\mathfrak{p}}\right)$ $\displaystyle=\left(\frac{\pm
1,-p}{2}\right)=\pm 1,$

where the second equation holds since
$F_{\mathfrak{p}}={\mathbb{Q}}_{2}(\sqrt{-d})={\mathbb{Q}}_{2}(\sqrt{p})$. ∎

Now we can prove Theorem 0.3 by using the above propositions.

###### Proof.

Let $L=\mathfrak{o}_{F}+\mathfrak{o}_{F}\sqrt{-1}$. Let $\mathfrak{p}$ be a
place of $F$ and $L_{\mathfrak{p}}$ be the $\mathfrak{p}$-adic completion of
$L$ inside $E_{\mathfrak{p}}=E\otimes_{F}F_{\mathfrak{p}}$. Recall
$T=R^{1}_{E/F}({\mathbb{G}}_{m,E})$ and $\mathbb{T}$ is the affine scheme
defined by the equation $x^{2}+y^{2}=1$, we have

$\displaystyle T(F)$ $\displaystyle=\\{\beta\in E^{*}:\ N_{E/F}(\beta)=1\\}$
$\displaystyle\mathbb{T}(\mathfrak{o}_{F_{\mathfrak{p}}})$
$\displaystyle=\\{\beta\in L_{\mathfrak{p}}^{\times}:\
N_{E_{\mathfrak{p}}/F_{\mathfrak{p}}}(\beta)=1\\}.$

And $L_{\infty}^{\times}=E_{\infty}^{*}=\mathbb{C}^{*}\times\mathbb{C}^{*}$.

Let $v\in\Omega_{E}$ and $v\mid 2$. Let $\mathfrak{p}$ be the unique place of
$F$ above $2$. By Proposition 1.3 and 1.8, one has

$\prod_{v\mid
2}\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},p}{v}\right)_{4}=1$

for any
$(x_{\mathfrak{p}},y_{\mathfrak{p}})\in\mathfrak{o}_{F_{\mathfrak{p}}}\times
o_{F_{\mathfrak{p}}}$ with $x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=1$.
Regard $\mathbb{T}(\mathfrak{o}_{F_{\mathfrak{p}}})$ as a subgroup of
$T(\mathbb{A}_{F})$, this implies that

$\lambda_{E}(\mathbb{T}(\mathfrak{o}_{F_{\mathfrak{p}}}))\subseteq
E^{*}N_{\Theta/E}(\mathbb{I}_{\Theta}).$

Let $v\in\Omega_{E}$ and $v\mid p$. Let $\mathfrak{p}$ be the unique place of
$F$ above $p$. By Proposition 1.2 and 1.7, one has

$\prod_{v\mid
p}\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},p}{v}\right)_{4}=1$

for any
$(x_{\mathfrak{p}},y_{\mathfrak{p}})\in\mathfrak{o}_{F_{\mathfrak{p}}}\times
o_{F_{\mathfrak{p}}}$ with $x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=1$. This
implies that

$\lambda_{E}(\mathbb{T}(\mathfrak{o}_{F_{\mathfrak{p}}}))\subseteq
E^{*}N_{\Theta/E}(\mathbb{I}_{\Theta}).$

And $\Theta/E$ is unramified over each place $v$ of $E$ except $v\mid 2p$.
Therefore the natural group homomorphism

$\widetilde{\lambda}_{E}:\nicefrac{{T(\mathbb{A}_{F})}}{{T(F)\prod_{\mathfrak{p}\leq\infty}\mathbb{T}(\mathfrak{o}_{F_{\mathfrak{p}}})}}\longrightarrow[\nicefrac{{\mathbb{I}_{E}}}{{E^{*}N_{\Theta/E}(\mathbb{I}_{\Theta})}}]\times[\nicefrac{{\mathbb{I}_{E}}}{{E^{*}\prod_{\mathfrak{p}\leq\infty}L_{\mathfrak{p}}^{\times}}}]$

is well-defined. By Proposition 1.1, we only need to show
$\widetilde{\lambda}_{E}$ is injective.

Let $u\in ker\widetilde{\lambda}_{E}$. Then there are $\beta\in E^{*}$ and
$i\in\prod_{\mathfrak{p}\leq\infty}L_{\mathfrak{p}}^{\times}$ with
$\lambda_{E}(u)=\beta i$. We have

$N_{E/F}(\beta)=N_{E/F}(i)^{-1}\in
F^{*}\cap(\prod_{\mathfrak{p}\leq\infty}\mathfrak{o}_{F_{\mathfrak{p}}}^{\times})=\\{\pm
1\\},$

since $F$ is an imaginary quadratic field and
$F\neq{\mathbb{Q}}(\sqrt{-1})\text{ or }{\mathbb{Q}}(\sqrt{-3})$.

If $N_{E/F}(\beta)\neq 1$, one obtains $N_{E/F}(\beta)=N_{E/F}(i)=-1$. Write
$i=(i_{v})_{v}\in\mathbb{I}_{E}$. Since $\Theta/E$ is unramified over each
place $v$ of $E$ except $v\mid 2p$, one concludes that
$\psi_{\Theta/E}(i_{v})$ is trivial for all primes $v\nmid 2p$, where $i_{v}$
is regarded as an idele whose $v$-component is $i_{v}$ and 1 otherwise.

(1) Suppose $(d,p)\in D_{1}\cup D_{2}$. Since $N_{E/F}(i_{v})=-1$ and
$i_{v}\in L_{\mathfrak{p}}^{\times}$, one gets

$\prod_{v\mid 2}\psi_{\Theta/E}(i_{v})=1\text{ and
}\psi_{\Theta/E}(i_{v^{\prime}})=-1$

by Proposition 1.2 and 1.3, where $\psi_{\Theta/E}:\mathbb{I}_{E}\rightarrow
Gal(\Theta/E)$ is the Artin map and $v^{\prime}$ is the unique place of $E$
above $p$. So

$\psi_{\Theta/E}(\beta i)=\psi_{\Theta/E}(i)=\prod_{v\mid
2}\psi_{\Theta/E}(i_{v})\cdot\psi_{\Theta/E}(i_{v^{\prime}})=-1.$

This contradicts to $u\in ker\widetilde{\lambda}_{E}$.

(2) Suppose $(d,p)\in D_{3}$. Since $N_{E/F}(i_{v})=-1$ and $i_{v}\in
L_{\mathfrak{p}}^{\times}$, one gets

$\psi_{\Theta/E}(i_{v^{\prime}})=-1\text{ and }\prod_{v\mid
p}\psi_{\Theta/E}(i_{v})=1$

by Proposition 1.7 and 1.8, where $v^{\prime}$ is the unique place of $E$
above $2$. So

$\psi_{\Theta/E}(\beta
i)=\psi_{\Theta/E}(i)=\psi_{\Theta/E}(i_{v^{\prime}})\cdot\prod_{v\mid
p}\psi_{\Theta/E}(i_{v})=-1.$

This contradicts to $u\in ker\widetilde{\lambda}_{E}$.

Therefore $N_{E/F}(\beta)=1$, one concludes that

$N_{E/F}(\beta)=N_{E/F}(i)=1\ \ \ \Rightarrow\ \ \ \beta\in T(F)\ \ \
\text{and}\ \ \
i\in\prod_{\mathfrak{p}\leq\infty}\mathbb{T}(\mathfrak{o}_{F_{\mathfrak{p}}}).$

So $\beta i\in
T(F)\prod_{\mathfrak{p}\leq\infty}\mathbb{T}(\mathfrak{o}_{F_{\mathfrak{p}}})$.
Then $\widetilde{\lambda}_{E}$ is injective. ∎

######  Lemma 1.9.

Let $F={\mathbb{Q}}(\sqrt{-2d})$ and $d\equiv 3\mod 4$. If $x_{\mathfrak{p}}$
and $y_{\mathfrak{p}}$ in $\mathfrak{o}_{F_{\mathfrak{p}}}$ satisfy
$x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=-1$, then the $4$-th Hilbert symbol

$\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},2}{v}\right)_{4}=-1$

where $v$ and $\mathfrak{p}$ are respectively the unique place of $E$ and $F$
above $2$.

###### Proof.

The Hilbert symbol

$\displaystyle\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},2}{v}\right)_{4}$
$\displaystyle=\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},-2d}{v}\right)_{4}\cdot\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},-d}{v}\right)^{-1}_{4}$
$\displaystyle=\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},-2d}{v}\right)_{4}\cdot
1=\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},\sqrt{-2d}}{v}\right)$
$\displaystyle=\left(\frac{-1,2d}{2}\right)=-1$

where the second equation holds by Lemma 1.6. ∎

Using a similar argument as in the proof of Theorem 0.3, the result follows
from Lemma 1.6 and 1.9.

######  Proposition 1.10.

Let $F={\mathbb{Q}}(\sqrt{-2d})$ and $d\equiv 3\mod 4$. Then the diophantine
equation $x^{2}+y^{2}=\alpha$ is solvable over $\mathfrak{o}_{F}$ if and only
if $\alpha$ satisfies the Artin condition of $\Theta$ and $H_{L}$, where
$\Theta=E(\sqrt[4]{2})$ and $H_{L}$ is the ring class field corresponding to
the order $L=\mathfrak{o}_{F}+\mathfrak{o}_{F}\sqrt{-1}$.

Now we use Proposition 1.10 to give an explicit example. Let
$F={\mathbb{Q}}(\sqrt{-6})$. We write
$N_{F/{\mathbb{Q}}}(\alpha)=2^{s_{1}}3^{s_{2}}p_{1}^{e_{1}}\cdots
p_{g}^{e_{g}}$ for any $\alpha=a+b\sqrt{-6}\text{ and }a,b\in{\mathbb{Z}}$.
Let $P(n)=\\{p_{1},\cdots,p_{g}\\}$. Denote $a=3^{s_{3}}a_{1}$ with $3\nmid
a_{1}$ and

$\displaystyle P_{1}=\\{p\in
P(n):\left(\frac{-1}{p}\right)=\left(\frac{-6}{p}\right)=1\text{ and
}\left(\frac{2}{p}\right)=-1\\}$ $\displaystyle P_{2}=\\{p\in
P(n):\left(\frac{-1}{p}\right)=-\left(\frac{-6}{p}\right)=1\text{ and
}\left(\frac{2}{p}\right)=-1\\}$ $\displaystyle P_{3}=\\{p\in
P(n):\left(\frac{-1}{p}\right)=\left(\frac{-6}{p}\right)=1\text{ and
}\left(\frac{2}{p}\right)_{4}=-1\\}.$

It’s easy to see that $e_{i}$ is even when $p_{i}\in P_{2}$.

######  Example 1.11.

Let $F={\mathbb{Q}}(\sqrt{-6})$ and $\alpha$ an integer in $F$ with the above
notation. Then $x^{2}+y^{2}=\alpha$ is solvable over $\mathfrak{o}_{F}$ if and
only if

(1) The equation $x^{2}+y^{2}=\alpha$ has integral solutions at every place of
$F$.

(2) $P_{1}\neq\emptyset$, or

$s_{1}/2+\sum_{p_{i}\in P_{2}}e_{i}/2+\sum_{p_{i}\in
P_{3}}e_{i}\equiv\begin{cases}0\mod 2\ \ \ &\text{if }a_{1}\equiv 1,3\mod 8\cr
1\mod 2\ \ \ &\text{if }a_{1}\equiv-1,-3\mod 8\end{cases}$

for $P_{1}=\emptyset$.

##  2\. The sum of two squares in real quadratic fields

Let $d>1$ be a square-free odd number and $F={\mathbb{Q}}(\sqrt{2d})$. Let
$\mathfrak{o}_{F}$ be the ring of integers of $F$, $\varepsilon_{F}$ the
fundamental unit of $\mathfrak{o}_{F}$ and $\varepsilon_{F}=a+b\sqrt{2d}\text{
with }a,b>0$. Let $E=F(\sqrt{-1})$. One takes the order
$L=\mathfrak{o}_{F}+\mathfrak{o}_{F}\sqrt{-1}$ inside $E$. In this section we
always assume that one of the equations $x^{2}-2dy^{2}=\pm 2$ is solvable over
${\mathbb{Z}}$ and we fix one solution $(x_{0},y_{0})$. Denote
$\omega=x_{0}+y_{0}\sqrt{2d}$ and $\eta=\omega^{2}/2$. Then
$\eta\in\mathfrak{o}_{F}^{\times}$ and $\eta=\varepsilon_{F}^{i_{0}}$ for some
$i_{0}\in{\mathbb{Z}}$. By the assumption, we have
$N_{F/{\mathbb{Q}}}(\varepsilon_{F})=1$ (see [Per], pp. 106-109).

######  Lemma 2.1.

Let $p\equiv\pm 3\mod 8$ and $p|d$. If one of the equations $x^{2}-2dy^{2}=\pm
2$ is solvable over ${\mathbb{Z}}$ with the notation as above, then $i_{0}$ is
odd and the $4$-th Hilbert symbol

$\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},p}{v}\right)_{4}=-1$

for $x_{\mathfrak{p}}$ and $y_{\mathfrak{p}}$ in
$\mathfrak{o}_{F_{\mathfrak{p}}}$ satisfy
$x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=\varepsilon_{F}$, where $v$ and
$\mathfrak{p}$ are respectively the unique prime in $E$ and $F$ above $p$.

###### Proof.

The Hilbert symbol

$\displaystyle\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},p}{v}\right)_{4}$
$\displaystyle=\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},2d}{v}\right)_{4}\cdot\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},2d/p}{v}\right)^{-1}_{4}$
$\displaystyle=\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},2d}{v}\right)_{4}\cdot
1=\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},\sqrt{2d}}{v}\right)$
$\displaystyle=\left(\frac{\varepsilon_{F},\sqrt{2d}}{\mathfrak{p}}\right)$

where the second equation holds since $E(\sqrt[4]{2d/p})/E$ is unramified over
$v$. However,

$\left(\frac{\eta,\sqrt{2d}}{\mathfrak{p}}\right)=\left(\frac{\omega^{2}/2,\sqrt{2d}}{\mathfrak{p}}\right)=\left(\frac{2,p}{p}\right)=-1.$

Then

$-1=\left(\frac{\eta,\sqrt{2d}}{\mathfrak{p}}\right)=\left(\frac{\varepsilon_{F},\sqrt{2d}}{\mathfrak{p}}\right)^{i_{0}}.$

Therefore $\left(\frac{\varepsilon_{F},\sqrt{2d}}{\mathfrak{p}}\right)=-1$ and
$i_{0}$ is odd. ∎

######  Lemma 2.2.

Let $p|d$, $p\equiv\pm 3\mod 8$. And one of the equations $x^{2}-2dy^{2}=\pm
2$ is solvable over ${\mathbb{Z}}$ with the notation as above. If
$x_{\mathfrak{p}}$ and $y_{\mathfrak{p}}$ in $\mathfrak{o}_{F_{\mathfrak{p}}}$
satisfy $x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=\varepsilon_{F}$, then

$\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},p}{v}\right)_{4}=1$

where $v$ and $\mathfrak{p}$ are respectively the unique place of $E$ and $F$
above $2$.

###### Proof.

Since $E_{v}=F_{\mathfrak{p}}(\sqrt{-1})$ is ramified over $F_{\mathfrak{p}}$,
there is a uniformizer $\pi_{F}$ in $F_{\mathfrak{p}}$ such that
$\left(\frac{\pi_{F},-1}{\mathfrak{p}}\right)=1$. We know
$x^{2}+y^{2}=\pi_{F}^{3}$ is solvable over $\mathfrak{o}_{F_{\mathfrak{p}}}$
and $x^{2}+y^{2}=\pi_{F}$ is not solvable over
$\mathfrak{o}_{F_{\mathfrak{p}}}$ (by Theorem 1 in [Ko]). Choose $(a,b)$ be
one solution of the equation $x^{2}+y^{2}=\pi_{F}^{3}$, then we have $a\text{
and }b\in\mathfrak{o}_{F_{\mathfrak{p}}}^{\times}$, otherwise we obtains
$x^{2}+y^{2}=\pi_{F}$ is solvable over $\mathfrak{o}_{F_{\mathfrak{p}}}$.

Recall $\omega=x_{0}+y_{0}\sqrt{2d}$ and $\eta=\omega^{2}/2$, here
$x_{0},y_{0}\in{\mathbb{Z}}$ satisfy $x_{0}^{2}-2dy_{0}^{2}=\pm 2$. Denote

$\epsilon=2^{-1}\omega\cdot(1+\sqrt{-1})\cdot(a+b\sqrt{-1})/(a-b\sqrt{-1}).$

We can see $N_{E_{v}/F_{\mathfrak{p}}}(\epsilon)=\eta$. And

$\displaystyle\epsilon$
$\displaystyle=2^{-1}\pi_{F}^{-3}\omega\cdot(1+\sqrt{-1})\cdot(a^{2}-b^{2}+2ab\sqrt{-1})$
$\displaystyle=2^{-1}\pi_{F}^{-3}\omega\cdot[(a^{2}-b^{2}-2ab)+(a^{2}-b^{2}+2ab)\sqrt{-1}]$
$\displaystyle=(\omega/2-b\omega(a+b)/\pi_{F}^{3})+(\omega/2-b\omega(b-a)/\pi_{F}^{3})\sqrt{-1}.$

In the following we will prove $a\not\equiv\pm b\mod 2$. If not then $a=\pm
b+2u$ for some $u\in\mathfrak{o}_{F_{\mathfrak{p}}}$. We have $(\pm
b+2u)^{2}+b^{2}=\pi_{F}^{3}.$ Note that
$b\in\mathfrak{o}_{F_{\mathfrak{p}}}^{\times}$, then

$2=ord_{\mathfrak{p}}(4u^{2}+2b^{2}\pm
4ub)=ord_{\mathfrak{p}}(\pi_{F}^{3})=3,$

a contradiction is derived. So $a\not\equiv b\mod 2$. Since $a,b$ are units
and the residue field of $\mathfrak{o}_{F_{\mathfrak{p}}}$ is
$\mathbb{F}_{2}$, we have

$a+b=\pi_{F}u_{1},a-b=\pi_{F}u_{2}$

with $u_{1},u_{2}\in\mathfrak{o}_{F_{\mathfrak{p}}}^{\times}$. Then

$\displaystyle\epsilon$
$\displaystyle=(\omega/2-bu_{1}\omega/\pi_{F}^{2})+\sqrt{-1}(\omega/2+bu_{2}\omega/\pi_{F}^{2})$
$\displaystyle=2^{-1}\omega(1-bu_{1}\frac{2}{\pi_{F}^{2}})+2^{-1}\omega(1+bu_{2}\frac{2}{\pi_{F}^{2}})\sqrt{-1}.$

We have
$\epsilon\in\mathfrak{o}_{F_{\mathfrak{p}}}+\mathfrak{o}_{F_{\mathfrak{p}}}\sqrt{-1}$
since $b,u_{1},u_{2}\in\mathfrak{o}_{F_{\mathfrak{p}}}^{\times}$ and the
residue field of $\mathfrak{o}_{F_{\mathfrak{p}}}$ is $\mathbb{F}_{2}$.

Let $\mathfrak{q}$ be the unique prime of ${\mathbb{Q}}(\sqrt{-1})$ above $2$.
Let $s\in{\mathbb{Z}}_{2}^{\times}$ such that $s^{2}=p/3$ or $-p/3$. For any
$\delta\in E_{v}^{*}$ satisfies $N_{E_{v}/F_{\mathfrak{p}}}(\delta)=\eta$, we
have

$\displaystyle\left(\frac{\delta,p}{v}\right)_{4}$
$\displaystyle=\left(\frac{\delta,3}{v}\right)_{4}\cdot\left(\frac{\delta,s}{v}\right)\cdot\left(\frac{\delta,\pm
1}{v}\right)_{4}=\left(\frac{\delta,3}{v}\right)_{4}\cdot\left(\frac{\eta,s}{\mathfrak{p}}\right)\cdot
1$
$\displaystyle=\left(\frac{\delta,3}{v}\right)_{4}\cdot\left(\frac{1,s}{p}\right)=\left(\frac{\delta,3}{v}\right)_{4}.$

Then

$\displaystyle\left(\frac{\epsilon,p}{v}\right)_{4}$
$\displaystyle=\left(\frac{2^{-1}\pi_{F}^{-3}\omega\cdot(1+\sqrt{-1})\cdot(a+b\sqrt{-1})^{2},3}{v}\right)_{4}$
$\displaystyle=\left(\frac{2^{-1},3}{v}\right)_{4}\cdot\left(\frac{\pi_{F}^{3},3}{v}\right)^{-1}_{4}\cdot\left(\frac{\omega,3}{v}\right)_{4}\cdot\left(\frac{1+\sqrt{-1},3}{v}\right)_{4}\cdot\left(\frac{a+b\sqrt{-1},3}{v}\right)$
$\displaystyle=\left(\frac{2^{-1},3}{\mathfrak{q}}\right)\cdot\left(\frac{\pi_{F}^{3},3}{v}\right)^{-1}_{4}\cdot\left(\frac{\pm
2,3}{\mathfrak{q}}\right)_{4}\cdot\left(\frac{1+\sqrt{-1},3}{\mathfrak{q}}\right)\cdot\left(\frac{\pi_{F}^{3},3}{\mathfrak{p}}\right)$
$\displaystyle=1\cdot\left(\frac{\pi_{F}^{3},3}{v}\right)^{-1}_{4}\cdot
1\cdot(-1)\cdot\left(\frac{\pi_{F},3}{\mathfrak{p}}\right)$

Since $\left(\frac{\pi_{F},-1}{\mathfrak{p}}\right)=1$ by the choice of
$\pi_{F}$ and $F_{\mathfrak{p}}(\sqrt{-3})/F_{\mathfrak{p}}$ is unramifed of
degree $2$, then one obtains

$\left(\frac{\pi_{F},3}{\mathfrak{p}}\right)=\left(\frac{\pi_{F},-3}{\mathfrak{p}}\right)=-1.$

Therefore

$\displaystyle\left(\frac{\epsilon,p}{v}\right)_{4}$
$\displaystyle=\left(\frac{\pi_{F}^{3},3}{v}\right)^{-1}_{4}=\left(\frac{N_{E_{v}/{\mathbb{Q}}_{2}}(a+b\sqrt{-1}),3}{\mathfrak{q}}\right)^{-1}_{4}.$

We can write $N_{E_{v}/{\mathbb{Q}}_{2}}(a+b\sqrt{-1})=2^{3}m$ with $m\equiv
1\text{ or }-3\mod 8$. Therefore

$\displaystyle\left(\frac{\epsilon,p}{v}\right)_{4}$
$\displaystyle=\left(\frac{2^{3}m,3}{\mathfrak{q}}\right)_{4}^{-1}=\left(\frac{m,3}{\mathfrak{q}}\right)_{4}^{-1}$
$\displaystyle=$ $\displaystyle\begin{cases}1&\text{ if }m\equiv 1\mod 8\\\
\left(\frac{-3,3}{\mathfrak{q}}\right)_{4}^{-1}=1&\text{ if }m\equiv-3\mod
8.\end{cases}$

Let $x_{\mathfrak{p}}$ and $y_{\mathfrak{p}}$ in
$\mathfrak{o}_{F_{\mathfrak{p}}}$ satisfy
$x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=\varepsilon_{F}$. Since
$\eta=\varepsilon_{F}^{i_{0}}$, we have

$\left(\frac{\epsilon,p}{v}\right)_{4}=\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},p}{v}\right)_{4}^{i_{0}}$

by Lemma 1.6. Since $i_{0}$ is odd by Lemma 2.1, one obtains

$\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},p}{v}\right)_{4}=1.$

∎

######  Corollary 2.3.

Let $F={\mathbb{Q}}(\sqrt{2d})$ and $p|d$, $p\equiv\pm 3\mod 8$. If one of the
equations $x^{2}-2dy^{2}=\pm 2$ is solvable over ${\mathbb{Z}}$, then
$x^{2}+y^{2}=\varepsilon_{F}$ is not solvable over $\mathfrak{o}_{F}$.

###### Proof.

Recall $\mathbb{X}_{\varepsilon_{F}}$ is the affine scheme over
$\mathfrak{o}_{F}$ defined by $x^{2}+y^{2}=\varepsilon_{F}$. Let
$\Theta=E(\sqrt[4]{p})$. One obtains

$\psi_{\Theta/E}(f_{E}[\prod_{\mathfrak{p}\leq\infty}(x_{\mathfrak{p}},y_{\mathfrak{p}})])=\left(\frac{x_{\mathfrak{p}^{\prime}}+y_{\mathfrak{p}^{\prime}}\sqrt{-1},p}{v}\right)_{4}=-1$

for any
$\prod_{\mathfrak{p}\leq\infty}(x_{\mathfrak{p}},y_{\mathfrak{p}})\in\prod_{\mathfrak{p}\leq\infty}\mathbb{X}_{\varepsilon_{F}}(\mathfrak{o}_{F_{\mathfrak{p}}})$
by Lemma 2.1 and 2.2, where $\psi_{\Theta/E}$ is the Artin map, $v$ and
$\mathfrak{p}^{\prime}$ are respectively the unique prime of $E$ and $F$ above
$p$ . The result follows from the class field theory. ∎

######  Remark.

In fact $x^{2}+y^{2}=\varepsilon_{F}$ has local integral solutions at every
place of $F$. The solvability is obvious if the place is not above $2$. Let
$\mathfrak{p}$ be the unique place of $F$ above 2. Then we only need to show
that it is solvable over $\mathfrak{o}_{F_{\mathfrak{p}}}$. Since
$\eta=\omega^{2}/2=\varepsilon_{F}^{i_{0}}$ with $i_{0}$ odd, we have

$x^{2}+y^{2}=\varepsilon_{F}\text{ is solvable over
}\mathfrak{o}_{F_{\mathfrak{p}}}\Leftrightarrow x^{2}+y^{2}=\eta\text{ is
solvable over }\mathfrak{o}_{F_{\mathfrak{p}}}.$

Note that $\omega=x_{0}+y_{0}\sqrt{2d}$ with $2\mid x_{0}$ and $y_{0}$ odd. It
is easy to verify that $x^{2}+y^{2}=\eta$ is solvable over
$\mathfrak{o}_{F_{\mathfrak{p}}}$ by Theorem 1 in [Ko].

######  Theorem 2.4.

Let $F={\mathbb{Q}}(\sqrt{2d})$. If $p|d$, $p\equiv\pm 3\mod 8$ and one of the
equations $x^{2}-2dy^{2}=\pm 2$ is solvable over ${\mathbb{Z}}$, then the
diophantine equation $x^{2}+y^{2}=\alpha$ is solvable over $\mathfrak{o}_{F}$
if and only if $\alpha$ satisfies the Artin condition of $\Theta$ and $H_{L}$,
where $\Theta=E(\sqrt[4]{p})$ and $H_{L}$ is the ring class field
corresponding to the order $L=\mathfrak{o}_{F}+\mathfrak{o}_{F}\sqrt{-1}$.

###### Proof.

Let $\mathfrak{p}$ be a place of $F$ and $L_{\mathfrak{p}}$ the
$\mathfrak{p}$-adic completion of $L$ inside
$E_{\mathfrak{p}}=E\otimes_{F}F_{\mathfrak{p}}$.

Let $v_{1}$ and $\mathfrak{p}_{1}$ be respectively the unique prime of $E$ and
$F$ above $2$. By Lemma 1.6, one has $\left(\frac{\xi,p}{v}\right)_{4}=1$ for
any $\xi\in L_{\mathfrak{p}_{1}}^{\times}$ with
$N_{E_{v_{1}}/F_{\mathfrak{p}_{1}}}(\xi)=1$. This implies that

$\lambda_{E}(\mathbb{T}(\mathfrak{o}_{F_{\mathfrak{p}_{1}}}))\subseteq
N_{\Theta_{\mathfrak{V}_{1}}/E_{v_{1}}}(\Theta_{\mathfrak{V}_{1}}^{*})$

where $\mathfrak{V}_{1}$ is the unique place of $\Theta$ above $v_{1}$.

Let $v_{2}$ and $\mathfrak{p}_{2}$ be respectively the unique prime of $E$ and
$F$ above $p$. By the similar computation in Lemma 2.1, one has
$\left(\frac{\xi,p}{v}\right)_{4}=1$ for any $\xi\in
L_{\mathfrak{p}_{2}}^{\times}$ with
$N_{E_{v_{2}}/F_{\mathfrak{p}_{2}}}(\xi)=1$. This implies that

$\lambda_{E}(\mathbb{T}(\mathfrak{o}_{F_{\mathfrak{p}_{2}}}))\subseteq
N_{\Theta_{v_{2}}/E_{v_{2}}}(\Theta_{v_{2}}^{*})$

where $\Theta_{v_{2}}=\Theta\otimes_{E}E_{v_{2}}$.

Since $\Theta/E$ is unramified over all primes except $v_{1}$ and $v_{2}$,
therefore the natural group homomorphism

$\widetilde{\lambda}_{E}:\nicefrac{{T(\mathbb{A}_{F})}}{{T(F)\prod_{\mathfrak{p}\leq\infty}\mathbb{T}(\mathfrak{o}_{F_{\mathfrak{p}}})}}\longrightarrow[\nicefrac{{\mathbb{I}_{E}}}{{E^{*}N_{\Theta/E}(\mathbb{I}_{\Theta})}}]\times[\nicefrac{{\mathbb{I}_{E}}}{{E^{*}\prod_{\mathfrak{p}\leq\infty}L_{\mathfrak{p}}^{\times}}}]$

is well-defined. By Proposition 1.1, we only need to show
$\widetilde{\lambda}_{E}$ is injective.

Let $u\in ker\widetilde{\lambda}_{E}$. Then there are $\beta\in E^{*}$ and
$i\in\prod_{\mathfrak{p}\leq\infty}L_{\mathfrak{p}}^{\times}$ with
$\widetilde{\lambda}_{E}(u)=\beta i$. Therefore

$N_{E/F}(\beta)=N_{E/F}(i)^{-1}\in
F^{*}\cap(\prod_{\mathfrak{p}\leq\infty}\mathfrak{o}_{F_{\mathfrak{p}}}^{\times})=\\{\pm\varepsilon_{F}^{n}|n\in{\mathbb{Z}}\\}.$

Since $N_{E/F}(\beta)$ is totally positive, we have
$N_{E/F}(\beta)=\varepsilon_{F}^{n}$ and $N_{E/F}(i)=\varepsilon_{F}^{-n}$.

Assume that $n$ is odd. Write $i=(i_{v})_{v}\in\mathbb{I}_{E}$. Since
$\Theta/E$ is unramified over all primes of $E$ except $v_{1}$ and $v_{2}$,
one has $\psi_{\Theta/E}(i_{v})$ is trivial for all primes $v\neq
v_{1},v_{2}$, where $i_{v}$ is regarded as an idele whose $v$-component is
$i_{v}$ and 1 otherwise. Since
$N_{E_{v}/F_{\mathfrak{p}}}(i_{v})=\varepsilon_{F}^{-n},$ one gets

$\psi_{\Theta/E}(\beta
i)=\psi_{\Theta/E}(i)=\psi_{\Theta/E}(i_{v_{1}})\psi_{\Theta/E}(i_{v_{2}})=1\cdot(-1)^{-n}=-1$

by Lemma 2.1 and 2.2, where $\psi_{\Theta/E}$ is the Artin map. This
contradicts to $u\in ker\widetilde{\lambda}_{E}$.

Therefore $n$ is even. Let

$\gamma=\beta\varepsilon_{F}^{n/2},j=i\varepsilon_{F}^{-n/2}.$

Then

$N_{E/F}(\gamma)=N_{E/F}(j)=1\ \ \ \Rightarrow\ \ \ \gamma\in T(F)\ \ \
\text{and}\ \ \
j\in\prod_{\mathfrak{p}\leq\infty}\mathbb{T}(\mathfrak{o}_{F_{\mathfrak{p}}}).$

So $\beta i=\gamma j\in
T(F)\prod_{\mathfrak{p}\leq\infty}\mathbb{T}(\mathfrak{o}_{F_{\mathfrak{p}}})$.
Then $\widetilde{\lambda}_{E}$ is injective. ∎

In the following we consider a special case. Let $d=p$ be a prime with
$p\equiv 3\mod 8$. Then the equation $x^{2}-2py^{2}=-2$ is solvable over
${\mathbb{Z}}$ (Corollary 2 in [Yo]).

######  Lemma 2.5.

Let $F={\mathbb{Q}}(\sqrt{2p})$ and $p\equiv 3\mod 8$. If $x_{\mathfrak{p}}$
and $y_{\mathfrak{p}}$ in $\mathfrak{o}_{F_{\mathfrak{p}}}$ satisfy
$x_{\mathfrak{p}}^{2}+y_{\mathfrak{p}}^{2}=\varepsilon_{F}$, then the $4$-th
Hilbert symbol

$\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},2}{v}\right)_{4}=-1$

where $v$ and $\mathfrak{p}$ are respectively the unique prime in $E$ and $F$
above $2$.

###### Proof.

The Hilbert symbol

$\displaystyle\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},2}{v}\right)_{4}$
$\displaystyle=\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},2p}{v}\right)_{4}\cdot\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},p}{v}\right)^{-1}_{4}$
$\displaystyle=\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},\sqrt{2p}}{v}\right)\cdot
1=\left(\frac{\varepsilon_{F},\sqrt{2p}}{\mathfrak{p}}\right)$

where the second equation holds by Lemma 2.2. Recall
$\omega=x_{0}+y_{0}\sqrt{2d}$ and $\eta=\omega^{2}/2$, here
$x_{0},y_{0}\in{\mathbb{Z}}$ satisfy $x_{0}^{2}-2py_{0}^{2}=-2$. And
$\eta=\varepsilon_{F}^{i_{0}}$ for some $i_{0}\in{\mathbb{Z}}$. By Lemma 2.1,
we have $i_{0}$ is odd. So we have

$\displaystyle\left(\frac{x_{\mathfrak{p}}+y_{\mathfrak{p}}\sqrt{-1},2}{v}\right)_{4}$
$\displaystyle=\left(\frac{\varepsilon_{F},\sqrt{2p}}{\mathfrak{p}}\right)=\left(\frac{\omega^{2}/2,\sqrt{2p}}{\mathfrak{p}}\right)$
$\displaystyle=\left(\frac{2,\sqrt{2p}}{\mathfrak{p}}\right)=\left(\frac{2,-2p}{2}\right)=-1.$

∎

Using a similar argument as in the proof of Theorem 2.4, the result follows
from Lemma 2.2 and 2.5.

######  Proposition 2.6.

Let $p$ be a prime and $F={\mathbb{Q}}(\sqrt{2p})$. If $p\equiv 3\mod 8$, then
the diophantine equation $x^{2}+y^{2}=\alpha$ is solvable over
$\mathfrak{o}_{F}$ if and only if $\alpha$ satisfies the Artin condition of
$\Theta$ and $H_{L}$, where $\Theta=E(\sqrt[4]{2})$ and $H_{L}$ is the ring
class field corresponding to the order
$L=\mathfrak{o}_{F}+\mathfrak{o}_{F}\sqrt{-1}$.

Now we use Theorem 2.6 to give an explicit example. Let
$F={\mathbb{Q}}(\sqrt{6})$. We write
$N_{F/{\mathbb{Q}}}(\alpha)=2^{s_{1}}3^{s_{2}}p_{1}^{e_{1}}\cdots
p_{g}^{e_{g}}$ for any $\alpha=a+b\sqrt{6}$, here $a,b\in{\mathbb{Z}}$. Let
$P(n)=\\{p_{1},\cdots,p_{g}\\}$. Denote $a=3^{s_{3}}a_{1}$ with $3\nmid a_{1}$
and

$\displaystyle P_{1}=\\{p\in
P(n):\left(\frac{-1}{p}\right)=\left(\frac{6}{p}\right)=1\text{ and
}\left(\frac{2}{p}\right)=-1\\}$ $\displaystyle P_{2}=\\{p\in
P(n):\left(\frac{-1}{p}\right)=-\left(\frac{6}{p}\right)=1\text{ and
}\left(\frac{2}{p}\right)=-1\\}$ $\displaystyle P_{3}=\\{p\in
P(n):\left(\frac{-1}{p}\right)=\left(\frac{6}{p}\right)=1\text{ and
}\left(\frac{2}{p}\right)_{4}=-1\\}.$

It’s easy to see that $e_{i}$ is even when $p_{i}\in P_{2}$.

######  Example 2.7.

Let $F={\mathbb{Q}}(\sqrt{6})$ and $\alpha$ an integer in $F$ with the above
notation. Then $x^{2}+y^{2}=\alpha$ is solvable over $\mathfrak{o}_{F}$ if and
only if

(1) The equation $x^{2}+y^{2}=\alpha$ has integral solutions at every place of
$F$.

(2) $P_{1}\neq\emptyset$, or

$s_{1}/2+\sum_{p_{i}\in P_{2}}e_{i}/2+\sum_{p_{i}\in
P_{3}}e_{i}\equiv\begin{cases}0\mod 2\ \ \ &\text{if }a_{1}\equiv\pm 1\mod
8\cr 1\mod 2\ \ \ &\text{if }a_{1}\equiv\pm 3\mod 8\end{cases}$

for $P_{1}=\emptyset$.

Acknowledgment  The work is supported by the Morningside Center of Mathematics
and NSFC, grant # 10901150 and # 10671104.

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