Re: Question on elliptic curves with same conductor and Diophantine triples

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

On Tue, Jun 04, 2024 at 12:46:05PM -0700, American Citizen wrote:
> Example 1
> 
> conductor                  elliptic curve Diophantine Triple
> 31141110   [0, -208619/396900, 0, -36542/33075, 7744/11025] [35/36, -27/35,
> 352/315]
> 31141110   [0, 868006/38025, 0, 1646161/38025, 559504/38025] [12/65, 143/60,
> 340/39]
> 
> Example 2
> 
> conductor                         elliptic curve Diophantine Triple
> 116867520  [0, 278088/4844401, 0, -11768632283520/23468221048801,
> 296284262400/23468221048801]  [40/9, 162/961, -756/5041]
> 116867520  [0, 952/9, 0, 6765440/2187,
> 11907174400/531441]                                      [22/9, 40/9, 124/9]
> 
> Please notice that in the second example the two triples contain a common
> 40/9 tuple.
> 
> 2. Given one triple [a,b,c] can we find another triple [d,e,f] associated
> with the same minimal elliptic curve? This process is essentially generating
> other elliptic curves with the same conductor but from other Diophantine
> triples.
> 
> Please bear in mind that the Diophantine relationship (1) has to satisfied
> in any triple we find.

One you can do is to use the invariants c4 and c6.
Two curves are isomorphic if ad only if they have same same covariants
c4, c6 up to to some multiplier u^4 resp. u^6.

? F=ellinit(E(a,b,c));
? F.c4
%19 = (16*b^2-16*c*b+16*c^2)*a^2+(-16*c*b^2-16*c^2*b)*a+16*c^2*b^2
? F.c6
%20 = (-64*b^3+96*c*b^2+96*c^2*b-64*c^3)*a^3+(96*c*b^3-384*c^2*b^2+96*c^3*b)*a^2+(96*c^2*b^3+96*c^3*b^2)*a-64*c^3*b^3

Now find you need to find a second triplet (aa,bb,cc) so that
ellinit(E(aa,bb,cc)) has the same invariant up to u^4 resp u^6 for some u.

For example for your second example.

? F1=ellinit(E(40/9, 162/961, -756/5041));
? F2=ellinit(E(22/9, 40/9, 124/9));
? ellisisom(F1,F2)
%61 = [369/2201,4708240/4844401,0,0]
? ispower(F1.c4/F2.c4,4,&u4);u4
%57 = 369/2201
? ispower(F1.c6/F2.c6,6,&u6);u6
%58 = 369/2201

Cheers,
Bill.