Re: another simple question

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

Using proper fractions from 1/2 to 11/12 (45 in all) and the negative 
values, I was able to find 215,433 elliptic curves with rational 
coefficients, containing the points [1/2, 5/8]

I found 8 interesting ones, composed of 10 dissimilar numbers (all 
positive) for the 5 coefficients:

> [3/4, 5/12, 9/10, 1/6, 7/8]
> [3/4, 7/12, 9/10, 5/6, 1/2]
> [3/4, 11/12, 7/10, 1/6, 5/8]
> [7/10, 2/3, 5/8, 11/12, 1/4]
> [9/10, 7/12, 5/8, 1/4, 2/3]
> [7/12, 3/8, 9/10, 5/6, 1/2]
> [11/12, 1/8, 9/10, 5/6, 2/3]
> [11/12, 5/8, 7/10, 1/6, 3/4]

On 4/18/24 05:39, Kurt Foster wrote:
> 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.