# An elementary question about Gaussian primes

Suppose we consider primes of the form $p = 1 \text { mod } 4$, so that $p = a^2 + b^2$, $a$ and $b$ being integers. Considering only the first quadrant, all $(a,b)$ pairs will be of the form (odd,even) or (even,odd).

Now if we consider $q = (a + n)^2 + (b + n)^2 = p + 2na + 2nb + 2n^2$, $n$ an integer, then $q = 1 \text { mod } 4$.

$1 \text { mod } 4$ as an arithmetic progression contains an infinite number of primes, and while every $q$ may not turn out to be a prime, one would still expect an infinite number of prime pairs $(p,q)$.

Has any work been done related to this? My original motivation was that this could give very close Gaussian primes, differing by a distance of just $n\sqrt2$, when $n$ is small.

Update: I came across a recent paper (Bounded gaps between Gaussian primes) where instead of $(a + n,b + n)$, the author considers $(a + n,b + m)$, with $(n,m)$ either $\text (odd,odd)$ or $\text (even,even)$, and then proves it for the case $(0,m)$, $m>=246$.

• Check out Bunyakovsky conjecture. – Wojowu Dec 26 '15 at 15:24
• This is a particular case of a natural conjecture that would generalize Dirichlet's theorem to the Gaussian integers, but this generalization is not known. I doubt this particular case is known. It is a famous open problem whether there are infinitely many primes in another arithmetic progression, $n+i$, or equivalently, infinitely many primes in the integers of the form $n^2+1$. – Douglas Zare Dec 26 '15 at 15:26
• I am pretty sure that Bunyakovsky conjecture is open for any specific nonlinear polynomial. And probably (less sure here) it is not even proved that there exists a nonlinear polynomial for which it holds. – Fedor Petrov Dec 26 '15 at 15:54
• Friedlander and Iwaniec's paper (arxiv.org/pdf/math/9811185.pdf) mentions that the problem in one variable is generally more difficult than that in two. So it's possible this problem is more tractable than Bunyakovsky's. – valar123 Dec 26 '15 at 18:05
• @valar123 this question is equivalent to the specific case of Bunyakovsky conjecture, is not it? – Fedor Petrov Dec 27 '15 at 14:54

An infinite number of primes on a ray of slope $45^{\circ}$ would not imply close Gaussian primes any more than ordinary primes $0^{\circ}.$ I didn't read the paper you link, but $246$ is the current record for gaps between ordinary primes. You might enjoy reading about Guassian moats.