if you look at the wolframalpha calculation page:

-13-26i - Wolfram|Alpha

it factorises -13-26i as

i×(1+2 i)×(2+3 i)×(3+2 i)

my question is why is theia necessary factor to take out? are the rules regarding this?

Thanks

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- Oct 4th 2011, 04:57 AMwalleyegaussian prime factorisation
if you look at the wolframalpha calculation page:

-13-26i - Wolfram|Alpha

it factorises -13-26i as

i×(1+2 i)×(2+3 i)×(3+2 i)

my question is why is the**i**a necessary factor to take out? are the rules regarding this?

Thanks - Oct 4th 2011, 05:40 AMTKHunnyRe: gaussian prime factorisation
Hint: (1-i) is NOT a Gaussian Prime. It can be factored: i(1+i)

- Oct 6th 2011, 08:59 PMroninproRe: gaussian prime factorisation
I don't think that this is correct. There is a characterisation of Gaussian primes that says that a Gaussian integer is prime if and only if one of the conditions is true:

- Its norm is prime (as an integer)
- It is a prime (as an integer) of the form

In this case, the norm of is , which is prime. So is prime.

The issue that should be brought up is that unique factorisation only holds*up to a unit*(i.e. invertible number). For example, normal integer factorisation follows this idea: or . In the second example, we have multiplied the factors by (which is invertible), but we do not consider it to be inherently different from the first. So in this case, only differs from by multiplying by the unit , which is no difference at all, as far as unique factorisation goes.

So to address original question, throwing in is analogous to repeatedly multiplying by with the usual integer factorisation. If it bothers you, you may let one of the terms absorb it. - Oct 6th 2011, 10:26 PMTKHunnyRe: gaussian prime factorisation
I'm making stuff up again?! Only explanation I could think of for the question presented. It's -i, anyway. Wow. Taking a break...

- Oct 6th 2011, 10:38 PMroninproRe: gaussian prime factorisation
No worries! Mistakes happen.