The number of Gaussian primes

Question

Let $n$ be a natural number $\ge 2$.

Let $\pi(n)$ be the number of primes $(p)$ for $p\le n$ and $\pi g(n)$ the number of Gausian primes $(z = x + yi )$ for $|x|$ and $|y| \le n$.

Can the ratio $\pi g(n)/\pi(n)^2$ has a limit for $n \to \infty$?

Answer 1

First, let's simplify the problem to the more natural variant where the bounds defining $\pi g(n)$ are $x^2 + y^2 \le n^2$. Since you mention $|x|,|y|$ I will assume you intend to include all conjugates of a given Gaussian prime in the count, but it's a distinction that doesn't make much difference to what follows.

Under this simplification, (let's call it) $\pi g'(n)$ is essentially $8$ times the number of primes congruent to $1 \pmod 4$ up to $n^2$, ignoring some insignificant terms: the $O(1)$ contribution from the prime $2$ and the $o(n)$ contribution from rational Gaussian primes like $7$. The latter is asymptotic to $n^2/(4 \log n)$ by the Prime Number Theorem for arithmetic progressions, so $\pi g'(n) \sim 2n^2/\log n$.

$\pi(n)$ is of course, asymptotic to $n/\log n$. So the ratio $\pi g'(n)/\pi(n)^2$ grows like $O(\log n)$ and thus it does not have a limit. Since your square bounding region is even larger than the circle, the same divergence holds for $\pi g(n)$.

It would be more interesting to eliminate the $\log n$ factor by considering something like $\pi g(n)/(n \pi(n))$. In that case I believe there are equidistribution results due to Hecke which would probably be enough to establish an asymptotic for the number of Gaussian primes inside the square bounding box (I expect it would be the same as the circle but scaled by a constant proportionate to the ratio of the areas, so $4/\pi$).