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
> ellj((1+sqrt(163)*I)/2)
> getting -262537412640768000
> <<
> However on the boundary of the fundamental domain, my calculation shows 
> only j-invariants ( positive) which are perfect cubes as
> Q(sqrt(-1)).....12^3
> Q(sqrt(-2))......20^3
> Q(sqrt(-4))......66^3
> Q(sqrt(-7))......255^3
> and the above almost integer. Are there any others on the Imaginary axis?
>

No. cm_j_invariants_and_orders(QQ) gives all 13 imaginry quadratic orders 
of class number 1, from which you can recover the 13 associated imaginary 
discriminants D.  Most of these are congruent to 1 mod 4 so the j-value is 
j((1+sqrt(d))/2) , only those which are 0 mod 4 are on the imaginary axis 
with values j(sqrt(D)/2) as in your list.

THere is a big theory of complex multiplcation behind these facts, but I 
don'y think that "gp in Sage" is an accurate sub                       ject 
line for discussion about that.

John Cremona

 

>
>
> On Mon, Aug 31, 2020 at 2:28 AM Dima Pasechnik <dim...@gmail.com> wrote:
>
>> On Sun, Aug 30, 2020 at 9:24 AM Dima Pasechnik <dim...@gmail.com> wrote:
>> >
>> > 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 ?
>>
>> 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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