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

["Ruud H.G. van Tol" <rvtol@isolution.nl>]

On 2024-11-04 01:03, hermann@stamm-wilbrandt.de wrote:
> On 2024-11-03 21:58, Ruud H.G. van Tol wrote:
> > On 2024-11-03 21:39, Bill Allombert wrote:
> > > On Sun, Nov 03, 2024 at 04:00:16PM +0100, 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]
> > > I suggest
> > >
> > > [w*z | w<-powers(I,3); z<-[c,conj(c)]]
> >
> > Another one:
> >
> > vecsort(concat(c*=[-1,1,-I,I], conj(c)),,8)
> >
> > (assuming that duplicates are undesired)
> Thanks for all who answered, so many different ways.
>
> My subject was misleading, what I was interested in was one gaussian 
> prime.
> And for that exactly 8 values exist, no duplicates possible.
>
> Since "powers(I,3)" as well "[-1,1,-I,I]" both have 11 characters,
> I like joining Bill's "powers" and Ruud's "*=" solution as shortest:
>
> ? c=4+I*7;
> ? concat(c*=powers(I,3),conj(c))
> [4 + 7*I, -7 + 4*I, -4 - 7*I, 7 - 4*I, 4 - 7*I, -7 - 4*I, -4 + 7*I, 7 
> + 4*I]
> ?

The noisy side is that it changes c.
Further it is good to use lexicals:

my(c=4+I*7, t=powers(I,3)*c); concat(t, conj(t))

-- Ruud