How many rational multiples of a revolution pass through square lattice points? Given the set $\mathbb{Z}^2$, if we draw a line passing through $(0,0)$ which is some fraction $a/b=r$ of a full revolution from $(1,0)$, how many rational solutions are there for $r$ which pass through non-zero points in $\mathbb{Z}^2$?
Clearly, any multiple of $1/8$ of a revolution passes through non-zero lattice points right next to the origin, so there are 8 angles that are trivial solutions, but are there any more?
 A: If I've understood your question right, how many rational angles (using the definition that a rational angle is one such that $\theta = a \cdot 360°, a \in \mathbb{Q}$) are there such that if you draw a line through the origin at that angle with the $x$-axis, at one point that line hits a lattice point?
The answer to this question is, unfortunately, no (unless you consider $\frac{3}{8}$, $\frac{5}{8}$ and $\frac{7}{8}$ other non-trivial solutions).
Here's a proof (it's probably wrong but whatever):
Let $\theta$ = $\frac{a}{b}$, with $a \in \mathbb{Z}$ and $b \in \mathbb{Z}_0$. The formula for a line through the origin given an angle in this format is $y = \tan(2\pi n)$. We want to know all points where that $\text{tan}(2\pi \theta)$ is an integer. You can find a value of $\tan(2\pi \theta)$ arbitrarily close to any integer (see this Math SE question for an explanation), but you'll never actually hit an integer. The only roots for $\tan(x)$ are of the form $\pi n$ with $n \in \mathbb{Z}$ which is unfortunately always going to be irrational.
The reason that $\frac{1}{8}$ (and its aforementioned cousins) do work is simply because $\frac{1}{8} \cdot 2\pi$ is $\frac{\pi}{4}$ and $\tan(\frac{\pi}{4})$ is one, so the formula for the line turns into $y = x$.
