Is there an effective way to decompose gaussian integers into prime factors?

Question

We define $\mathbb{Z}[i] := \{a + bi \mid a, b \in \mathbb{Z}\}, i = \sqrt{-1},$ which is an euclidean ring together with $N: \mathbb{Z}[i] \to \mathbb{N}_0, z \mapsto z\bar{z}=a^2+b^2$ for $z=a+bi$. Being an euclidean ring means also to be a factorial ring so that each element $z \in \mathbb{Z}[i]$ can be decomposed into a product of prime elements.

Given an exercise to decompose $4 + 12i$ into prime factors in $\mathbb{Z}[i]$ I spent a lot of time just bruteforcing the combinations. Then I got $$4+12i = 4(1 + 3i) = 2^2(-1 + i)(1-2i) = (1+i)^2(1-i)^2(-1+i)(1-2i),$$ which also was the intended solution. There is also a theorem that states that if $N(z)$ is a prime in $\mathbb{Z}$, then $z$ is irreducible in $\mathbb{Z}[i]$, hence it is a prime element in $\mathbb{Z}[i]$. That verifies that all factors on the right side of the solution are prime.

As bruteforce is never effective, scalable and good for ones' karma I asked myself if there is a more effective way to compute this decomposition.

Are you aware of any procedure that brings some insights into the decomposition and makes it more effective?

Answer 1

Dario Alpern's Gaussian integer factorisation calculator has, like many other programs there, an explanation of the mathematics behind it.

Suppose you have the prime factorisation of the norm $N(z)$; primes of the form $4n+3$ will always appear in pairs. Each integer prime factor or pair – with multiplicity – corresponds to a Gaussian prime factor of $z$ as follows:

• The ramified prime $2$ corresponds to $1+i$.
• The split prime $4n+1$ can be written as $m^2+n^2$ in a unique way with $m>n$. Then either $m+ni$ or $m-ni$ is a factor.
• The inert prime pair $(4n+3)^2$ corresponds to the Gaussian and integer prime $4n+3$.

As an example, for factoring $4+12i$ by hand we notice that it has content (GCD of coefficients) $2^2$, so $$4+12i=(-1)(1+i)^4(1+3i)$$ $1+3i$ has norm $2×5$. So there is another $1+i$ factor and the quotient is a prime: $$4+12i=(-1)(1+i)^5(2+i)$$ Note that the factorisation is only unique up to units.