Re: question on solutions of rational versus integer ternary quadratics

The reason for all those solutions appearing in both Z and Q is due to a 
silly C++ programming error for the lgcd(a,b) routine when either or 
both are zero.

The correct return value for (0,a) or (a,0) is abs(a)

For (0,0) it is 1

After fixing my lgcd(a,b) routine, now the solutions correctly correspond

I apologize for this subtle error in the gcd portion of my C++ code.

Randall

On 7/14/25 13:11, American Citizen 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.
>