Bill Allombert on Sun, 07 Jul 2024 11:37:36 +0200
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Re: Question: trying to locate other Diophantine triples from certain elliptic curves
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- To: pari-users@pari.math.u-bordeaux.fr
- Subject: Re: Question: trying to locate other Diophantine triples from certain elliptic curves
- From: Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>
- Date: Sun, 7 Jul 2024 11:37:32 +0200
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On Sat, Jul 06, 2024 at 06:57:34PM -0700, American Citizen wrote:
> Hello:
>
> I am working with elliptic curves associated with a Diophantine triple
> [a,b,c] such that a*b+1, a*c+1, and b*c+1 are all rational squares:
>
> The equation of this curve is
>
> (1) E_triple(a,b,c) = [0,(a*b+a*c+b*c),0,(a*b*c)*(a+b+c),(a*b*c)^2]
>
> in Weierstrass format.
>
> For a selected triple [a,b,c] = t = [5/4, 5/36, 32/9], the curve becomes
>
> (2) E(5/4,5/36,32/9) = E(t) = [0, 6625/1296, 0, 2225/729, 2500/6561]
>
> The minimal model of this curve is
>
> (3) M(t) = [1, 0, 0, -593988, 172568592] with conductor 510450
>
> GP-Pari ellisomat(E_triple) command gives four isogenous classes for (2)
> E(t) in [0,0,0,a,b] Weierstrass format
>
> K_1 = [-28511425/5038848, 149142030625/29386561536]
> K_2 = [-453599425/5038848, 9660698574625/29386561536]
> K_3 = [3360575/5038848, 464053326625/29386561536]
> K_4 = [-3935425/314928, -4396529375/459165024]
>
> Using the expression for the c4 invariant to (1) above I came up with
>
> (4) F.c4 = (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
>
> And using brute force searching for b,c, by solving for (4) = E(t).c4
> invariant = 28511425/104976 and the fact that the "a" variable
> is a quadratic in F.c4, I came up with a list of over 164 triples which have
> the same c4 value for the elliptic curve E(t).
>
> My apparently mistaken idea was that I could find isomorphic curves by
> matching the c4 value. This apparently is naive and too simplistic.
By definition, ellisomat never return isomorphic curves.
If you want to create isomorphic curves, you can use F=
ellchangecurve(E,[u,r,s,t])
for any [u,r,s,t] with u invertible.
then we have
1) F.c4/E.c4 = u^4
2) F.c6/E.c6 = u^6
and conversely, if it exists u such that 1) and 2) are true, then
E and F are isomorphic.
In particular, if [a,b,c] is a triplet [u*a,u*b,u*c] is also a triplet
with isomorphic curve.
Cheers,
Bill