On Sun, Aug 30, 2020 at 5:50 AM Surendran Karippadath
<
kksin...@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 ?
>
> 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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