Re: Question: trying to locate other Diophantine triples from certain elliptic curves

[John Cremona <john.cremona@gmail.com>]

(briefly)

To get a point on the curve you need the product of the three factors to be a square. The stronger condition that each factor separately is a square is simply the condition that the point is double another point.  So getting one point is enough: if the separate factors are not squares, double the point!

I agree entirely with Bill.

John

On Tue, 9 Jul 2024, 17:24 American Citizen, <website.reader3@gmail.com> wrote:

We're only 1/2 way done here, and the condition for [a,b,c] a Diophantine triple needs to be applied.

On 7/9/24 08:47, Bill Allombert wrote:

So if (x,y) is a point on E(a,b,c) then 
y/(c*b+x) , y/(c*a+x) and y/(b*a+r) is another triple.

An example:

E_triple(a,b,c)  = [0,(a*b+a*c+b*c),0,(a*b*c)*(a+b+c),(a*b*c)^2]
E=ellinit(E_triple(5/4,5/36,32/9));
ellrank(E)
F=ellchangecurve(E,[1, -235/54,0,0])
E_triple(475/36,-19/60,-50/171)==F[1..5]

The values [475/36, -19/60, -50/171] work, but unfortunately they are NOT a Diophantine triple.

I actually took my ellpool() algorithm, asked for all points on E up to height 100, there were 1,067 such points

Then I found all triples, using the formula above (actually the 3rd term is y/(b*a+x) ) and checked to see if

any DIophantine triples might be found. Yes, but only 2 were found.

Point on E  - triple

[[0, -50/81], [-5/4, -5/36, -32/9]]
[[0, 50/81], [5/4, 5/36, 32/9]]

were the only 2 triples found, but they create exactly the same E_triple curve we started with, so nothing

was gained here.

This is only the first part of my two-part question, the other was working with the isogenous curves in short Weierstrass [0,0,0,A,B] format to see if they can be transformed to the E_triple(a,b,c) format with [a,b,c] a Diophantine triple.