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

[American Citizen <website.reader3@gmail.com>]

I appreciate all your effort, Bill and John

Somehow, someway, something got sideways.

The Diophantine condition is that (a*b+1), (a*c+1) and (b*c+1) are 
squares.  I don't understand how this got changed to (a*b), (a*c) and 
(b*c) squares.

Can we start over again, with this understanding?

Thanks


On 7/9/24 10:42, Bill Allombert wrote:
> On Tue, Jul 09, 2024 at 05:33:53PM +0100, John Cremona wrote:
> > (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!
> To fix my 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));
> R=ellrank(E);
> P=ellmul(E,R[4][1],2)
> F=ellchangecurve(E,[1,1169363/27075,0,0]);
> [A,B,C]=[P[2]/x|x<-nfroots(,elldivpol(F,2))]
> E_triple(A,B,C)==F[1..5]
> [issquare(A*B),issquare(B*C),issquare(A*C)]
>
> ? 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));
> ? R=ellrank(E);
> ? P=ellmul(E,R[4][1],2)
> %4 = [1169363/27075,-83444186119/277789500]
> ? F=ellchangecurve(E,[1,1169363/27075,0,0]);
> ? [A,B,C]=[P[2]/x|x<-nfroots(,elldivpol(F,2))]
> %6 = [42422057/6727140,14766269/2147380,44462068/6418485]
> ? E_triple(A,B,C)==F[1..5]
> %7 = 1
> ? [issquare(A*B),issquare(B*C),issquare(A*C)]
> %8 = [1,1,1]
>
> Cheers,
> Bill.