Thank you all for the temporary solution to my problem arising from an
ambitious effort to understand Table 12.12 in David Cox's book Primes of
the form x^2+ny^2. As Prof Cremona has stated the existence of *only* four
perfect cubes on the imaginary axis is to be discussed under an
appropriate
On Wednesday, September 2, 2020 at 5:00:07 PM UTC+1 kks wrote:
> Yes, I knew the point regarding
> >>
> ndeed, there are 9 imaginary quadratic extensions of Q for which one
> gets integer j-invariant, one of them
> Q[sqrt(-163)], but as 163 mod 4 = 3, one has to compute its j-invariant as
>
Yes, I knew the point regarding
>>
ndeed, there are 9 imaginary quadratic extensions of Q for which one
gets integer j-invariant, one of them
Q[sqrt(-163)], but as 163 mod 4 = 3, one has to compute its j-invariant as
ellj((1+sqrt(163)*I)/2)
getting -262537412640768000
<<
However on the boundary of
On Sun, Aug 30, 2020 at 9:24 AM Dima Pasechnik wrote:
>
> On Sun, Aug 30, 2020 at 5:50 AM Surendran Karippadath
> wrote:
> > I evaluated the j-invariant in Pari/gp In SageMathCell
> > ? \p 50
> > ? ellj(sqrt(163.0)*I)
> > %1 = 68925893036109279891085639286944512.0163739
>
> Sage has
On Sun, Aug 30, 2020 at 5:50 AM Surendran Karippadath
wrote:
> I evaluated the j-invariant in Pari/gp In SageMathCell
> ? \p 50
> ? ellj(sqrt(163.0)*I)
> %1 = 68925893036109279891085639286944512.0163739
Sage has this function too (it calls Pari, so that's not an
independent
Hi,
I evaluated the j-invariant in Pari/gp In SageMathCell
? \p 50
? ellj(sqrt(163.0)*I)
%1 = 68925893036109279891085639286944512.0163739
Furthermore the Cube-root of the j-invariant I obtained
? (ellj(sqrt(163.0)*I))^(1/3)
%2 = 410009702400.00077461269365317226812447191214259043
