Is there a simpler way to get all 8 gaussian integers with same norml2()

hermann on Sun, 03 Nov 2024 16:00:21 +0100

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Is there a simpler way to get all 8 gaussian integers with same norml2() ?

To: pari-users@pari.math.u-bordeaux.fr
Subject: Is there a simpler way to get all 8 gaussian integers with same norml2() ?
From: hermann@stamm-wilbrandt.de
Date: Sun, 03 Nov 2024 16:00:16 +0100
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? id(c)=c;
? neg(x)=-x;
? flip(c)=I*conj(c);
? c=7+4*I;vecsort([f(g(h(c)))|f<-[id,conj];g<-[id,neg];h<-[id,flip]])

[-7 - 4*I, -7 + 4*I, -4 - 7*I, -4 + 7*I, 4 - 7*I, 4 + 7*I, 7 - 4*I, 7 +4*I]

?

Regards,

Hermann.

Follow-Ups:
- Re: Is there a simpler way to get all 8 gaussian integers with same norml2() ?
  - From: John Cremona <john.cremona@gmail.com>
- Re: Is there a simpler way to get all 8 gaussian integers with same norml2() ?
  - From: Loïc Grenié <loic.grenie@gmail.com>
- Re: Is there a simpler way to get all 8 gaussian integers with same norml2() ?
  - From: Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>

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