2+i is not a multiple of 2-i. They are independent prime divisors of 5. 1+i and 1-i _are_ associates; each is a multiple of the other. Rich ---- Quoting Marc LeBrun <mlb@well.com>:
="Mike Stay" <metaweta@gmail.com>
-Marc LeBrun <mlb@well.com> wrote: But x=1-i has no "positive" Gaussian integer divisors at all. That's because it's (-i)(1+i).
Doh, you're right!
I stand corrected, so the sum of first-quadrant divisors of 1-i is 2+i.
Sigh, I probably should've stuck with 5.
So, are not 1, 2+i and 5 the "first quadrant" Gaussian integer divisors of 5, with sum 8+i?
In contrast with the "positive" Gaussian integer divisors of 5 (as I recalled from Conway&Guy), which also include 2-i, and thus sum to 10?
I can rotate 2+i by a unit vector to get 2-i, but I'm durned if I can see how to factor that vector into first-quadrant Gaussian integers.
But then, I've been wrong before...
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