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

 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:
```
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.
>>
>
>
```
```

```