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Let $z = x+ iy \in \mathbb{Z}[i]$ and let $a+ib \in \mathbb{Z}[i]$ with $a^2 + b^2 \equiv 1 \mod{4}$. What is the probability that $a+ib$ divides $x + iy$ in $\mathbb{Z}[i]$? This question would resolve my answer to the probability that two gaussian integers are coprime as I asked here:

Probability that $x \equiv 3 \pmod{4}$

Since I know from an answer on that thread that the probability that $p \equiv 1 \mod{4}$ is $1/3 = 1/|\left(\mathbb{Z} / 4\mathbb{Z}\right)^{\times}|$, I could also just ask the probability that any two arbitrary gaussian integers are divisible and then multiply that answer by $1/3$.

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Samuel Reid
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  • Shouldn't the probability be zero, unless you put a bound on $a$ and $b$? – Zev Chonoles Mar 30 '13 at 21:20
  • @ZevChonoles: I'm not sure, care to explain? – Samuel Reid Mar 30 '13 at 21:22
  • Given any fixed $x+iy\in\mathbb{Z}[i]$, it only has finitely many divisors, while $\mathbb{Z}[i]$ is infinite. – Zev Chonoles Mar 30 '13 at 21:22
  • @ZevChonoles In the ring of integers ${\cal O}_K$ of a number field $K$, an integer is divisible by $n\in{\cal O}$ iff $x\equiv0$ in ${\cal O}/(n)$, so we may speak of $n\mid x$ happening with probability $|{\cal O}/(n)|^{-1}=N(n)^{-1}$ (which would be $1/(a^2+b^2)$ here, regardless of what $a^2+b^2$ is mod 4). My answer to the linked question covers this ground, but could benefit from more detail. – anon Mar 30 '13 at 21:59
  • @anon: I think the problem was that it's not sufficiently clear from the question that $a+\mathrm ib$ is fixed and $x+\mathrm iy$ is "asymptotically uniformly" random; one could read the question the other way around (and I suspect that Zev did). – joriki Mar 30 '13 at 22:04

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