Let $L$ be the line $y = \gamma x$ through the origin with irrational slope. That is: $\gamma \notin \Bbb{Q}$. I know that $L \cap \Bbb{Z^2} = \{(0,0)\}$. And that non-zero integer values for $x$ will come arbitrarily close to a lattice point. This post tells how to find the integer $x$ that comes closest to a lattice point. That is $| \lfloor \gamma x\rceil - \gamma x|$ is at a minimum for some range of $x$. What I want to find is the following:
For fixed irrational $\gamma$, $\epsilon > 0$, and some range $[A,B]\subset \Bbb{Z}$ Find all integers $x \in [A,B]$ such that $| \lfloor \gamma x\rceil - \gamma x| < \epsilon$.
In other words I'd like an algorithm to find all close lattice points to the line $L$ not just the shortest.
I know how to use lattices (black magic) to find the closest vector.
Reduce the lattice: $$\left(\begin{array}{ccc} 1 & 0 & -B \\ 0 & 1 & -\lfloor \gamma B\rceil \end{array} \right)$$
You will get a shortest vector that looks like: $(\star, t,\star )$. That $t$ is your answer. $| \lfloor \gamma t\rceil -\gamma t|$ seems to be a minimum for $t \in [0,B]$ A high powered mathematician showed me how to do this but he appears not to have time (or patience) to explain why this works, or how to generalize it.
Can anyone explain to me the gist of why this works and give any way to produce all such "close vectors" over some range for some $\epsilon$?
Thanks in advance.