Size and residue of the Rz candidate norm integer

Put $\delta =1+\omega$ and $q=1+\sqrt{2}$. Direct multiplication gives ${\delta }^2=q\sqrt{2}\omega$, ${\delta }^4=2q^{2}i$, and ${\delta }^{-1}=(1-\omega +{\omega }^2-{\omega }^3)\mathrm{/}2$. Since $q^{-1}=\sqrt{2}-1\in O$, these identities give $D=O[{\delta }^{-1}]$. Choose the least integer $h\ge 0$ such that $u=x\mathrm{/}{\delta }^h$ with $x=a{\omega }^3+b{\omega }^2+c\omega +d\in O$. Multiplying $x$ by the displayed inverse of $\delta$ shows that $x\mathrm{/}\delta \in O$ exactly when $a+b+c+d$ is even: each numerator coefficient is a signed sum of the four coefficients, and all four sums have the same parity. Thus if $h>0$, this sum is odd.

If $h=0$, the two disk conditions give $\mathrm{\mid }x{\mathrm{\mid }}^2+\mathrm{\mid }{x}^{\bullet }{\mathrm{\mid }}^2=2(a^2+b^2+c^2+d^2)\le 2$. Consequently $x=0$ or $x=\pm {\omega }^j$ for $0\le j\le 3$. If $h=1$, the same identity and $\mathrm{\mid }\delta {\mathrm{\mid }}^2+\mathrm{\mid }{\delta }^{\bullet }{\mathrm{\mid }}^2=4$ give $a^2+b^2+c^2+d^2\le 2$; its parity is odd by minimality, so it equals $1$. Then $x$ is again a root of unity, contradicting $\mathrm{\mid }{u}^{\bullet }{\mathrm{\mid }}^2=1\mathrm{/}(2-\sqrt{2})>1$. Thus the cases $h\le 1$ give only $u=0$ (where $n=1$) and the eight roots of unity (where $n=0$).

For $h\ge 2$, multiplication gives $x^{†}x=A+B\sqrt{2}$, where $A=a^2+b^2+c^2+d^2$ and $B=cd+bc+ab-da$. The odd sum of coefficients makes $A$ odd; also $B\equiv (a+c)(b+d)\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}2)$ since the two factors have opposite parity. Write $({\delta }^{†}\delta {)}^h=(2+\sqrt{2}{)}^h=C+D\sqrt{2}$. Both $C,D$ are even, since $(2+\sqrt{2}{)}^2=6+4\sqrt{2}$. It follows that $\xi =(X+Y\sqrt{2})\mathrm{/}(2+\sqrt{2}{)}^h$, with $X=C-A$ odd and $Y=D-B$ even. Taking real conjugates gives ${\xi }^{\bullet }\xi =(X^2-2Y^2)\mathrm{/}{2}^h$. Its numerator is $1(\mathrm{m}\mathrm{o}\mathrm{d}8)$, hence odd, so this denominator is already minimal. This proves the residue assertion; nonnegativity follows from $0\le \xi ,{\xi }^{\bullet }\le 1$.

For nonzero $u$, write $u=v\mathrm{/}(\sqrt{2}{)}^k$, $v\in O$. Then $\xi =(2^k-v^{†}v)\mathrm{/}{2}^k$, and $2^k-v^{†}v\in \mathbb{Z}[\sqrt{2}]$ by the same multiplication as in step 3.1. Therefore $2^{2k}{\xi }^{\bullet }\xi$ is an integer, whence $\mathrm{\ell }\le 2k$. The inequality $0\le {\xi }^{\bullet }\xi \le 1$ now yields $n\le 2^{\mathrm{\ell }}\le 4^k$.