Re: interesting discovery about elliptic curve [0,0,0, 393129,0]

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

Bill and John:

Thank you for your explanation and example.

Randall

On 10/24/24 05:13, Bill Allombert wrote:
> On Thu, Oct 24, 2024 at 08:56:27AM +0100, John Cremona wrote:
> > This is not an explanation, but the condition that x is a square is
> > equivalent to the point (x,y) being in the image of the 2-isogeny from
> > [0,0,0,-4n^2,0].    Calling the curves E and E' and looking at how descent
> > by 2-isogeny works (e.g. in my book, other books are available!), roughly
> > speaking the rank of E comes partly  from E/phi'(E') and part from the
> > image under phi of E'/phi(E).  (Here phi:E --> E') and phi' is the dual.)
> >   Saying that "all" the points have square x-coordinates is therefore saying
> > that E/phi'(E') is trivial and all the points are coming from phi'(E').
> >
> > But why that should be, apart from chance, I don't know.
> The conclusion is that this can happen with positive probability
> for curves having 2-torsion.
>
> To give an example of John's explanation:
> Take
> E = ellinit([393129,0]);
> E is 2-isogenous to
> F = ellinit([-1572516,0]);
>
> Both curves are of rank r=3 and have same BSD values.
>
> ? ellbsd(F)/ellbsd(E)
> %24 = 8.0000000000000000000000000000000000000
> is equal to 2^r so (assuming they have the same Tate-Shafarevich group),
> this implies that the image of F(Q) by phi^ is [2]E(Q),
> so every x is a square.
>
> Cheers,
> Bill.