Re: another simple question

[Kurt Foster <drsardonicus@earthlink.net>]

On Apr 17, 2024, at 6:55 PM, American Citizen wrote:

> To all:
>
> I am not sure how my post got misinterpreted, maybe this is my  
> fault??? I would be interested in your comments.
>
> When I said that many elliptic curves can be found with the point  
> [1/2, 5/8], I meant curves with the Weierstrass format  
> [a1,a2,a3,a4,a6] with all coefficients a1..a6 in Q, not Z.
<snip>
You can use the kernel I posted to find as many curves with rational  
a1, a2, a3, a4, a6 as you like.  I prefer making the coefficient a1 of  
x*y equal to 0, but to each his own.

> ? x=1/2;y=5/8;v=[y^2,x*y,y,-x^3,-x^2,-x,-1];
> ? w=v*denominator(v);
> ? M=matrix(7,7,i,j,if(i==1,w[j],0));
> ? K=matkerint(M)
> %4 =
> [0 0 0 0 0 -4]
>
> [0 0 0 0 -2 1]
>
> [1 1 1 -1 1 0]
>
> [1 -1 1 1 0 0]
>
> [0 1 0 1 0 -1]
>
> [1 1 -1 0 0 0]
>
> [0 0 1 -1 0 -1]

If you want the coefficient of y^2 to be 1 and also the coefficient of  
x*y equal to 0, you have to use -1/4*K[,6] -1/8*K[,5] which is

[1, 0, -1/8, 0, 1/4, 0, 1/4]~

For the coefficient of v[4] = -x^3 also to be 1, add any a*K[,1] +  
b*K[,2] + c*K[,3] + d*K[,4] with a - b + c + d = 1.

For example,

-1/4*K[,6]-1/8*K[,5]+K[,1] = [1, 0, 7/8, 1, 1/4, 1, 1/4]~

gives the curve

y^2 +7/8*y = x^3 + 1/4*x^2 + x + 1/4

E=ellinit([0,1/4,7/8,1,1/4])

which contains the given point.