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Note: A Gaussian integer is a complex number $a+bi$ where $a,b \in \mathbb Z$.
A Gaussian integer is prime if $a,b \neq 0$, then $a^2 + b^2$ is an ordinary prime, or if either $a=0$ or $b=0$, then $b$ or $a$ is a ordinary prime congruent to $3 \pmod 4$.

How could one go about showing the claim is true?
My first thought is something with Diriclet's Theorem on Primes in Arithmetic Progressions: Let ( a ) and ( d ) be two coprime positive integers (gcd(a, d) = 1). Then the arithmetic sequence:

a, a + d, a + 2d, a + 3d, ...

contains infinitely many prime numbers. That is, there exist infinitely many primes ( p ) satisfying:

$p \equiv a \pmod{d}$.

But I'm unsure on how to event construct a proof like this if even possible. Would appreciate any feedback or direction!

Thomas Andrews
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PeakyBlaze7788
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1 Answers1

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It's a known open problem that's highly likely to be yes, but it's unproven yet. One probabilistic argument for it: the chance that a number around size $n^2$ is prime ~ $1/log(n^2)$. We have n options there, so roughly can estimate expected count of primes as $n/2*log(n)$, the higher n, the higher expected count. So it's unlikely to find counterexample for big n(and all small ones were verified). There are different similar open problems and some that were proved to be true(for example for $n^2+k^4$ form). It has been discussed on this site already, you can check this one Finding Gaussian Primes along Lines in $\mathbb Z[i]$

alex st
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