When does $\|z^2\|=\|z\|^2$ Let $k \in \mathbb{Z}$ and consider the field extension $K := \mathbb{Q}[\sqrt{k}]$. Define a norm on $K$ given by $\|p+q\sqrt{k}\| := \sqrt{p^2+q^2}$. For any $z \in K$, I was interested to know when $\|z^2\| = \|z\|^2$.
It can be shown that $z=p+q\sqrt{k}$ satisfies $\|z^2\|=\|z\|^2$ if, and only if, 
$$q^2(k+1)(2p^2+(k-1)q^2)=0.$$
Clearly, $q=0$ means that $z \in \mathbb{Q}$ and $\sqrt{p^4}=\left(\sqrt{p^2}\right)^{\! 2}$ for all $p \in \mathbb{Q}$. When $k=-1$ we have $\mathbb{Q}[\operatorname{i}]$ which is a subset of $\mathbb{C}$, and we already know that the modulus satisfies $|wz|=|w||z|$.
My Questions: 


*

*What is the significance of the condition $2p^2+(k-1)q^2=0$? 

*Does it give any information about the norm $\|p+q\sqrt{k}\| = \sqrt{p^2+q^2}$?

*For a fixed $k$, what is so special about the $p+q\sqrt{k}$ which satisfy the condition?

 A: Let's focus on solutions of $2p^2+(k-1)q^2=0$ in $\mathbb{Q}$ for a fixed $k\in\mathbb{Z}$.


*

*If $k>1$, there is clearly only one solution $p=q=0$.

*If $k=1$, then $p=0$, $q\in\mathbb{Q}$ (even though I assume you don't want to consider cases where $k$ is a square).

*If $k<1$, then as before $p=0$, $q=0$ is a trivial solution. Assume $p^2+q^2>0$.
$$2p^2=(1-k)q^2$$
If $k$ is odd, then $k=1-2t$ for $t\ge1$. We get $$p^2=\frac{1-k}{2}q^2$$ $$p^2 = tq^2$$ Clearly $p\ne0\ne q$ so $t=\frac{p^2}{q^2}$ and therefore $\sqrt{t}=\left|\frac{p}{q}\right|\in\mathbb{Q}$. It's known that a square root of a (positive) integer is either an integer or irrational. Therefore $\sqrt{t}=s^2$, $s\in\mathbb{N}$. So $k=1-2s^2$ and $p=\pm sq$. If odd $k<1$ is not of this form, the only solution is the trivial one.
If $k$ is even, let $p=\frac{u_1}{u_2}$ and $q=\frac{v_1}{v_2}$ where $u_1,u_2,v_1,v_2\in\mathbb{Z}-\{0\}$ and $u_1,u_2$ and $v_1,v_2$ are relatively prime. Then
$$2\frac{u_1^2}{u_2^2}=(1-k)\frac{v_1^2}{v_2^2}$$
$$2u_1^2 v_2^2=(1-k)v_1^2 u_2^2$$
$$2u^2=(1-k)v^2$$
for $u,v\in\mathbb{Z}$, $u\ne 0\ne v$.
Since $k$ is even, multiplicity of the prime factor $2$ of LHS is odd and  multiplicity of the prime factor $2$ of RHS is even. Therefore the last equation does not have any solutions.

Let's summarize all results (not considering squares for $k$):


*

*$k>1$, $k$ is not a square

*

*$(p,0)$ for $p\in\mathbb{Q}$


*$k=-1$

*

*$(p,q)$ for $p,q\in\mathbb{Q}$


*$k<-1$, $k=1-2s^2$ for $s\in\mathbb{N}$

*

*$(p,0)$ for $p\in\mathbb{Q}$ 

*$(\pm sq,q)$ for $q\in\mathbb{Q}$


*$k<-1$, $k\ne 1-2s^2$ for $s\in\mathbb{N}$

*

*$(p,0)$ for $p\in\mathbb{Q}$ 


