Fiber of $-1$ of norm sum in the ring of integers Let $K$ be a real quadratic number field of discriminant $D>0$ with $\mathcal O_K$ being its ring of integers.
I'm interested in the map
$$ \varphi: \mathcal O_K \times \mathcal O_K \to \mathbb Z, \quad (\alpha,\beta) \mapsto N(\alpha)+N(\beta). $$
Is it surjective? Or can the fiber of $-1$ even be empty? If $\mathcal O_K$ has units of negative norm (which is for example the case when $D$ is prime) the latter is of course impossibe.
 A: Thm. The map $\varphi$ is surjective.
I will break the proof into three parts.
I. Preliminaries.
Write $K=\mathbb{Q}(\sqrt{d}),d>1$. One can assume that $d$ is squarefree.
First of all,  note that $\varphi(\alpha,\beta)=\det\pmatrix{\alpha & \beta \cr -\beta^* & \alpha^*}$
, where $*$ is the unique non trivial $\mathbb{Q}$-automorphism of $K$, so the image of $\varphi$ is a multiplicative monoid.
Note now that $0,=\varphi(0,0), 2=\varphi(1,1)$ and that $\varphi(m\alpha,m\beta)=m^2\varphi(\alpha,\beta)$ for all $m\in\mathbb{Z}$, so all in all, we are reduced to prove that the equation $$\varphi(\alpha,\beta)=a, \ \ \alpha,\beta\in\mathcal{O}_K$$ has at least a solution when $a$ is odd and squarefree.
Set $\omega=\sqrt{d}$ if $d\not\equiv  1 [4]$, and $\omega=\dfrac{-1+\sqrt{d}}{2}$ if $d=1+4k$, so that $\mathcal{O}_K=\mathbb{Z}[\omega]$.
If $\alpha=x+y\omega$ and $\beta=z+t\omega$, we have  $$\varphi(\alpha,\beta)=\left\lbrace\begin{array}{ccc}x^2+z^2-d(y^2+t^2)& if & d \not\equiv 1 [4]\cr  x^2-xy-ky^2+z^2-zt-kt^2 & if &  d=1+4k\end{array}\right.$$
Note that in the case $1+4k$, we also have $$\varphi(\alpha,\beta)=(x-\dfrac{y}{2})^2-dy^2+(z-\dfrac{t}{2})^2-dt^2  \ (*) $$
In particular, in both cases, $\varphi$ is an indefinite (since $d>0$) integral quadratic form in $4$ variables, which is non degenerate over $\mathbb{Q}$.
Now we have the following theorem.
Thm. (Eichler and al., 50's) Let $a\in\mathbb{Z}$, and let $q$ be an indefinite integral quadratic form in at least four variables which is non degenerate over $\mathbb{Q}$. Then $q$ represents $a$ over $\mathbb{Z}$ if and only if it represents $a$ over $\mathbb{Z}_p$ for all prime $p$.
Hence, we are reduced to prove the following facts:
Let  $a\in\mathbb{Z}$ be an odd squarefree integer, and let $p$ be a prime number . Then:
(a) the equation $x^2+z^2-d(y^2+t^2)=a, x,y,z,t\in\mathbb{Z}_p$ has at least one solution
(b) the equation $x^2-xy-ky^2+z^2-zt-kt^2=a, x,y,z,t\in\mathbb{Z}_p$ has at least one solution.
II. $p$ is odd.
I will make free use of the following well-known fact:
Fact. Let $a\in\mathbb{Z}_p$. If $a$ is a  nonzero square mod $p$, $a$ is a square in $\mathbb{Z}_p$.
Lemma 1. Let $a\in\mathbb{Z}_p^\times$.  Then the equation $a=u^2+v^2, u,v\in\mathbb{Z}_p$ has a solution.
Proof. It is known that this equation has a solution modulo $p$ (the set of elements of the form $\bar{v}^2,\bar{v}\in\mathbb{F}_p$ has $\dfrac{p-1}{2}+1=\dfrac{p+1}{2}$ elements, and same for the set of elements of the form $\bar{a}-\bar{u}^2$, so they cannot be disjoint).
Let $u_0,v_0\in\mathbb{Z}$ such that $a\equiv u_0^2+v_0^2 \ [p]$.
Since $a$ is not divisible by $p$ (this is a unit of $\mathbb{Z}_p$ !) , one may assume that $u_0\not \equiv 0 \ [p]$.
Hence $a-v_0^2$ is a nonzero square modulo $p$, hence a square in $\mathbb{Z}_p$ and we are done.
Proposition 2. Equations (a) and (b) have solutions in $\mathbb{Z}_p$.
Proof. Note that $2$ is a unit of $\mathbb{Z}_p$, so equation (b) has a solution in $\mathbb{Z}_p$ as soon as equation (a) has, since the substitution $(x,y,z,t)\longleftrightarrow (x-\dfrac{y}{2},y, z-\dfrac{t}{2},t)$ is bijective.
Hence it is enough to consider equation (a).
First case : $p\nmid a$.
Apply Lemma 1 to get $u,v\in\mathbb{Z}_p$ such that $u^2+v^2=a$. Then setting $x=u, y=0, z=v, t=0$ yields the desired solution for equation (a).
Second case : $p\mid a$.
First subcase: $p\mid d$, so $d=pd'$ where $p\nmid d'$ (since $d$ is squarefree). Write $a=pa'$, where $p\nmid a'$ (since $a'$ is squarefree). Then $d',a'$ are units of $\mathbb{Z}_p$, and so $-(d')^{-1}a'\in\mathbb{Z}_p^\times$ By Lemma 1, there exists $u,v\in\mathbb{Z}_p$ such that $u^2+v^2=-(d')^{-1}a'$. Then $-d(u^2+v^2)=pa'=a$.
Now set $x=0,y=u,z=0,t=v$.
Second subcase: $p\nmid d$. This time, $d$ is a unit of $\mathbb{Z}_p$, and so is $-d^{-1}(a-1)$. By Lemma 1, there exists $u,v\in\mathbb{Z}_p$ such that $u^2+v^2=-d^{-1}(a-1)$.
Then set $x=1,y=u,z=0,t=v$.
III. $p=2$.
I will make free use of the following well-known fact:
Fact. Let $a\in\mathbb{Z}_2$. If $a\equiv 1 \ [8]$, then $a$ is a square in $\mathbb{Z}_2$.
Proposition 3. Equations (a) and (b) have solutions in $\mathbb{Z}_2$.
Proof. Since $a$ is odd, we have $a\equiv 1,3,5,7 \ [8]$.
Using a CAS, one may check that equations (a) and (b) have a solution modulo $8$ of the form $(1,y_0,z_0,t_0)$ for all possible values of $a, d$ or $k$ modulo $8$ (of course, we have to exclude $d\equiv 0,4 \ [8]$ since $d$ is squarefree and stick to the values of $a$ listed above).
Hence $a- ( z_0^2-d(y_0^2+t_0^2))$ in case of Equation (a), and $a-(-y_0-ky_0^2+z_0^2-z_0t_0-kt_0^2)$ in case of equation (b), are congruent to $1$ modulo $8$, hence are square in $\mathbb{Z}_2$, and we are done.
