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

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

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.