Kevin Acres on Wed, 09 Nov 2016 12:06:55 +0100

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Re: Issquare and rational Intmods

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:


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.



At 08:04 PM 9/11/2016, John Cremona wrote:
On 8 November 2016 at 20:50, Kevin Acres <> 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?


> Sent from my iPhone
>> On 9 Nov 2016, at 7:33 AM, Bill Allombert <> 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.