Re: question on solutions of rational versus integer ternary quadratics

William Jagy <[email protected]>
Newsgroups gmane.comp.mathematics.pari.user
Message-ID <CACvx6ecHf10-2jGnefjRGR2KiujFaEYaA70DS_yOp4603v=RZA@mail.gmail.com>
I sent you an email a few hours ago, pointing out that you had introduced a
mistaken coefficient (1779 rather than the correct 1739) , at least in the
top line of one of your output files

On Mon, Jul 14, 2025 at 1:12 PM American Citizen <[email protected]>
wrote:

> Hello:
>
> I am working with the following conic and 2 ternary quadratics
>
> conic =  [261/5, -3481/50, 261/5, -1739/100, -1739/100, -1]
> T9Q([x,y,z]) = 261/5*x^2 + (-3481/50*y - 1739/100*z)*x + (261/5*y^2 -
> 1739/100*z*y - z^2)
> T9Z([x,y,z]) = 5220*x^2 + (-6962*y - 1739*z)*x + (5220*y^2 - 1739*z*y -
> 100*z^2)
>
> Over the range of integers x,y,z, for -10,000 <= x,y,z <= 10,000 I found
> 2367 solutions for T9Q, but only 1076 for T9Z. However the 1291
> solutions from T9Q plugged into T9Z do work.
>
> Why is T9Q missing 1,291 solutions?
>
> Randall
>
> btw: These ternary quadratic solutions collapse to just 113 points on
> the conic, but the [0,0] point has to be removed.
>
> Another btw note: I had special rational points on conics, it takes 2
> rational points to successfully recover all rationals on the conic,
> using the point-slope method to parameterize, and the reason is simple,
> for y - y1 = m(x-x1) where [x1,y1] is known and m is the slope, if you
> set m=0 for slope zero, the equation collapses to y - y1 = 0, and that
> only recovers your original [x1,y1] point and misses the 2nd y point
> which IS on the conic. I had to use 2 points and only found this out
> after laboring on this for at least 3 days wondering why my rational
> lattice points were missing some rationals.
>
>
>
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