Exponent growth of the Rz scaled grid

Write $q=1+\sqrt{2}$ and $C=(1+\sqrt{2})\mathrm{/}2$. For any integer $m$, the unit (using $q^{-1}=\sqrt{2}-1$ for negative $m$) $\lambda =q^m\in \mathbb{Z}[\sqrt{2}]$ satisfies $\mathrm{\mid }{\lambda }^{\bullet }\mathrm{\mid }={\lambda }^{-1}$. For arbitrary real $x,y$, solve the two real linear equations $\lambda (a+\sqrt{2}b)=x$ and ${\lambda }^{\bullet }(a-\sqrt{2}b)=y$ for real $a,b$ and round each to an integer by [F3]. The resulting $\alpha \in \mathbb{Z}[\sqrt{2}]$ satisfies $\mathrm{\mid }\alpha -x\mathrm{\mid }\le C\lambda$ and $\mathrm{\mid }{\alpha }^{\bullet }-y\mathrm{\mid }\le C{\lambda }^{-1}$. Applying this independently to real and imaginary coordinates gives, for arbitrary complex centers $z,w$, a $v\in O$ such that $\mathrm{\mid }v-z\mathrm{\mid }\le \sqrt{2}C\lambda$ and $\mathrm{\mid }{v}^{\bullet }-w\mathrm{\mid }\le \sqrt{2}C{\lambda }^{-1}$.

Set $s=2^{k\mathrm{/}2}$. Choose the integer $m$ so that $4\sqrt{2}C\mathrm{/}(sR)\le \lambda <4q\sqrt{2}C\mathrm{/}(sR)$. If $s^{2}rR\ge 128$, then $32C^{2}q=56+40\sqrt{2}<128$ implies $\sqrt{2}C\lambda <sr\mathrm{/}4$, while $\sqrt{2}C{\lambda }^{-1}\le sR\mathrm{/}4$. Apply step 1.1 to centers of disks in $sA$ and $(\sqrt{2}{)}^{k\bullet }B$; the latter disk has radius $sR$, since $\mathrm{\mid }(\sqrt{2}{)}^{k\bullet }\mathrm{\mid }=s$. Both $v$ and $v+\lambda$ lie in the first disk: their errors are bounded by $sr\mathrm{/}4$ and $sr\mathrm{/}4+\lambda <sr\mathrm{/}4+sr\mathrm{/}(4\sqrt{2}C)<sr$. Their bullet images likewise lie in the second disk, since the corresponding errors are at most $sR\mathrm{/}4$ and $sR\mathrm{/}4+sR\mathrm{/}(4\sqrt{2}C)<sR$. Dividing by $(\sqrt{2}{)}^k$ gives two distinct simultaneous grid points.

Put $\delta ={\epsilon }^2\mathrm{/}2$. The disk centered at $(1-\delta \mathrm{/}2)e^{-i\theta \mathrm{/}2}$ with radius $r=\delta \mathrm{/}2={\epsilon }^2\mathrm{/}4$ lies in the unit disk and in the half plane $\mathrm{Re}(e^{i\theta \mathrm{/}2}u)\ge 1-\delta$ by the triangle inequality. Take this as the disk in $A$ and the full unit disk as $B$. Since $2^{K}r\ge 128$, step 2.1 gives two candidates of exponent at most $K$. Every candidate is nonzero because its displayed real part is positive; [F2] therefore gives nonnegative exponents and finitely many candidates through each fixed exponent. In particular $k_2\le K$.

If $u\mathrm{\ne }v$ are grid points of exponent at most $k_2$, then for every integer $\mathrm{\ell }\ge 0$ the $2^{\mathrm{\ell }}+1$ points $w_h=(h\mathrm{/}{2}^{\mathrm{\ell }})u+(1-h\mathrm{/}{2}^{\mathrm{\ell }})v$, $0\le h\le 2^{\mathrm{\ell }}$, are distinct. They have exponent at most $k_2+2\mathrm{\ell }$, and both their original and bullet images lie in the required convex regions, because rational coefficients are fixed by $\bullet$. Thus the grid is infinite. Together with finite layers this proves the existence of its stated enumeration, and taking $\mathrm{\ell }=\lceil {\log }_{2}j\rceil$ gives the exponent bound.