Re: bnfisintnorm(bnfinit(x^2+3),7^5) misses integral solution

[Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>]

On Wed, Nov 18, 2020 at 02:22:47PM +0200, Georgi Guninski wrote:
> bnfisintnorm() appears to miss integral solutions, is this
> expected?
> 
> ? bnfisintnorm(bnfinit(x^2+3),7^5)
> %25 = [-149/2*x - 25/2, 133/2*x + 119/2, -147/2*x + 49/2, 147/2*x +
> 49/2, -133/2*x + 119/2, 149/2*x - 25/2]
> ? norm(Mod(2+x,x^2+3)^5)-7^5
> %26 = 0
> ? Mod(2+x,x^2+3)^5
> %27 = Mod(-31*x - 118, x^2 + 3)

Hello Georgi,

Note that the documentation says "modulo units of positive norm".
Let u the third root of unity:
? u=Mod(-1/2*x+1/2,x^2+3);
? norm(u)
%2 = 1
? (-149/2*x - 25/2)*u
%3 = Mod(-31*x-118,x^2+3)

So the list is complete modulo units of positive norm.
Cheers,
Bill.