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

[hermann@stamm-wilbrandt.de]

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]
?

Regards,

Hermann.