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

[Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>]

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.