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
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
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.