On Sun, Aug 30, 2020 at 9:24 AM Dima Pasechnik <dimp...@gmail.com> wrote:
>
> On Sun, Aug 30, 2020 at 5:50 AM Surendran Karippadath
> <kksinfin...@gmail.com> wrote:
> >      I evaluated the j-invariant in Pari/gp In SageMathCell
> > ? \p 50
> > ? ellj(sqrt(163.0)*I)
> > %1 = 68925893036109279891085639286944512.000000000163739
>
> Sage has this function too (it calls Pari, so that's not an
> independent confirmation that this number is (not) an integer:
> sage: elliptic_j(sqrt(163)*I,prec=500)
> 6.89258930361092798910856392869445120000000001637386442092346075751855217523117650690239250072955532985645916831850173541132959651401661828116253839333e34
>
> The output in Pari is a bit easier to read:
>
> ? \p 500
> ? ellj(sqrt(163)*I)
> %4 = 
> 68925893036109279891085639286944512.00000000016373864420923460757518552[...]
>
> Is it one of these "almost integers" (unless it's a bug, and this
> number must be an integer, I don't know  - number theorists, please
> step forward!), such as
> ? \p 500
> ? exp(sqrt(163)*Pi)
> %3 = 
> 262537412640768743.99999999999925007259719818568887935385633733699086270[...]
>
> Or a bug in Pari/GP ?

If I read the discussion  after Cor. 42 in
http://people.maths.ox.ac.uk/greenbj/papers/ramanujanconstant.pdf
right, this is not an integer.
Indeed, 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


>
>
> >
> > Furthermore the Cube-root of the j-invariant I obtained
> > ? (ellj(sqrt(163.0)*I))^(1/3)
> > %2 = 410009702400.00077461269365317226812447191214259043
>
> the closest integer to (ellj(sqrt(163.0)*I))^(1/3) is 410009702400,
> and so you can check
> that its cube is quite far from ellj(sqrt(163)*I)
>
>
>
> >
> > Is it possible to check in Sage with High Precision if the values are 
> > Integers.
> >
> > Thanking you in advance
> >
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