Re: Is there a simpler way to get all 8 gaussian integers with same norm

John Cremona on Sun, 03 Nov 2024 17:01:55 +0100

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

To: hermann@stamm-wilbrandt.de
Subject: Re: Is there a simpler way to get all 8 gaussian integers with same norml2() ?
From: John Cremona <john.cremona@gmail.com>
Date: Sun, 3 Nov 2024 16:01:41 +0000
Cc: Pari Users <pari-users@pari.math.u-bordeaux.fr>
Delivery-date: Sun, 03 Nov 2024 17:01:56 +0100
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In-reply-to: <7a5e5d01059136b4f9b57a0360286185@stamm-wilbrandt.de>
References: <7a5e5d01059136b4f9b57a0360286185@stamm-wilbrandt.de>

You could use qfminim to get all solutions up to sign.

John

On Sun, 3 Nov 2024, 15:00 , <hermann@stamm-wilbrandt.de> wrote:

? 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.

References:
- Is there a simpler way to get all 8 gaussian integers with same norml2() ?
  - From: hermann@stamm-wilbrandt.de

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