TITLE: Primes and the Unit circle.
QUESTION [5 upvotes]: Consider the "prime spiral" $f(z) = \sqrt{z}\exp(2\pi i \sqrt{z})$, for integer $z$. It has been shown that the intersections of $f$ with some quadratic curves contain a significantly disproportionate number of primes. Let $P = \{\exp(2\pi i \sqrt{p}) \;:\ p \;prime \}$, the set of unitized prime spiral coordinates.
1) Is $P$ equal to the circle group or is it a proper subset? (note that $P$ is countable)
2) What is the complement of $P$ in the circle group?
3) Is the complement a group?
4) Is the circle group equal to the closure of $P$?

REPLY [3 votes]: As a complement to Gerry's answer, another straightforward way of showing that $P$ can't be the circle group is to note that $P$ isn't even a group; there is no $n$ with $\sqrt{n}\equiv\sqrt{2}+\sqrt{3}\pmod 1$.  For if there were, then there would be some $m$ with $\sqrt{n}=\sqrt{2}+\sqrt{3}+m$; but then $n=(2+3+m^2)+2m\sqrt{2}+2m\sqrt{3}+2\sqrt{6}$ ; and $\sqrt{2}, \sqrt{3}$ and $\sqrt{6}$ can be easily shown to be linearly independent over the integers.