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I'm trying to factorise the Gaussian integer $z =11 - 3i$ into Gaussian primes. Taking the Euclidean norm on $z$ is $\nu (z) = 130$ which factorises into $2 \times 5 \times 13$ and so I'm assuming I want to find Gaussian primes $a, b, c$ such that $\nu(a) = 2,\ \nu(b) =5,\ \nu(c) = 13$ ? But I can't see how to go about doing this? Am I on the right track or is there a set method to factorise into gaussian primes?

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    Do you know a classification of the Gaussian primes?2012-06-03
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    You need to write those number as a sum of 2 squares. I think trial and error will easily give you the answer.2012-06-03
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    which numbers? $2, 5, 13$?2012-06-03

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Hint: Try $(1+i)(1-i)$, $(2+i)(2-i)$, and $(3+2i)(3-2i)$.

Useful Fact: a number is a sum of two squares if and only if each prime factor of that number that is equal to $3\pmod{4}$ occurs with even exponent.

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    And if those did not work, you might have to give $(1+2i), (1-2i), (2+3i), (2-3i) $ a go.2012-06-03
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    Ah, ok so I understand that if a prime, p = 1 (mod 4) then you can write it as $qq^*$ where they're both Gaussian primes. And if prime, p = 3 (mod 4) then it's a Gaussian prime.2012-06-03
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    But what I don't understand is that after we've split $\nu(z)$ into a factorisation of Gaussian primes. Then how does this relate to $z$?2012-06-03
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    @morphism: remember that $(11-3i)(11+3i)=130$. This means that you would have to try one of each conjugate pair of the factors of $130$ to get $11-3i$. The product of the others from the pairs would give $11+3i$. For example: $$(1+i)(2+1)(3+2i)=-3+11i$$ If we try the conjugates, we get $$(1-i)(2-i)(3-2i)=-3-11i$$ Multiply the latter by $i$ to get $$(1+i)(2-i)(3-2i)=11-3i$$2012-06-03
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    Ah! I understand how they're linked now, thank you2012-06-03