Re: Issquare and rational Intmods

[Kevin Acres <research@research-systems.com>]

Hi John,

Assuming that I have 'a' and 'm' where 'a' is an arbitrary x 
coordinate and 'm' a gradient. I use a similar process to 
issquare(Mod(-3*a^2,m),&Y) to help me find solutions for a 
progression of two further coordinates 'b' & 'c' such that:

        k=((a*b+b*c+a*c)/(2*m))^2-a*b*c

where c = m^2-a-b and k+a^3, k+b^3 and k+c^3 are all rational squares.

You can probably see that I'm plotting linearly dependent rational 
points on a related series of elliptic curves.

Currently integer 'a' and rational 'm' is working fine, it would just 
be nice to cope with rational 'a' as well.

Whilst I'm here, a note to Bill.  ellheight used to be able to take a 
vector of points but it seems that this feature has disappeared in 
2.9.0. I'm not sure is this is by design or just an oversight.

Regards,

Kevin.





At 08:04 PM 9/11/2016, John Cremona wrote:
> On 8 November 2016 at 20:50, Kevin Acres 
> <research@research-systems.com> wrote:
> > Hi Bill,
> >
> > For example I need to calculate something like issquare(Mod(-1727/4, 37/6))
>
> What do you expect Mod(-1727/4, 37/6) to mean?
>
> John
>
> >
> > Sent from my iPhone
> >
> >> On 9 Nov 2016, at 7:33 AM, Bill Allombert 
> <Bill.Allombert@math.u-bordeaux.fr> wrote:
> >>
> >>> On Wed, Nov 09, 2016 at 07:01:25AM +1100, Kevin Acres wrote:
> >>> I'm needing to perform an issquare(Mod(x,y),&s) where both x and y may
> >>> be rational and the square root is returned in s.
> >>>
> >>> Any help on how to do this correctly would be gratefully received.
> >>
> >> You need to explain mathematicaly what you want: please write down the
> >> equation 's' must satisfy.
> >>
> >> Mod(x,y) is only defined when y is an integer (or a polynomial).
> >>
> >> Cheers,
> >> Bill.
> >>
> >
> >