Solving a pair of ternary quadratic form equations

Question

Let $Q_1(x_0, x_1, x_2), Q_2(x_0, x_1, x_2) \in \mathbb{Z}[x_0, x_1, x_2]$ be two primitive, non-singular ternary quadratic forms (possibly indefinite). Suppose we want to solve the simultaneous equations

$$\displaystyle u_1 = Q_1(x_0, x_1, x_2), u_2 = Q_2(x_0, x_1, x_2), u_1, u_2 \in \mathbb{Z}, \gcd(x_0, x_1, x_2) = 1.$$

Then according to this paper, a theorem of Mordell states that the solutions can be found as follows: first consider the conic defined by

$$\displaystyle C_{u_1,u_2} : Q_{u_1, u_2}(x_0, x_1, x_2) = u_1 Q_2(x_0, x_1, x_2) - u_2 Q_1(x_0, x_1, x_2) = 0.$$

If there is a solution at all, this conic must have a rational point, which can be detected using the local-to-global principle. Having found a rational point, one can then use the "slope/intersect" trick to produce a parametrization. That is, draw lines emanating from a global point which must exist by Hasse's principle: this line will intersect the conic at another rational point and this sweeps out all of the rational points on the conic. This produces binary quadratic forms $f_0(u,v), f_1(u,v), f_2(u,v)$ so that the conic $C_{u_1,u_2}$ is exactly parametrized by $z_i = f_i(u,v), i = 0,1,2$.

One then inserts these quadratic forms into $Q_1, Q_2$, producing binary quartic forms, and then solve the two Thue equations given by $u_i = Q_i(f_0(u,v), f_1(u,v), f_2(u,v))$.

What I am not quite sure is the number of parametrizations. That is, there could be inequivalent triples of quadratic forms $(f_0, f_1, f_2)$ which parametrize the points on $C_{u_1, u_2}$. How do we bound the number of distinct parametrizations?

Answer 1

Part of this is finding the primitive integer null vectors of an indefinite ternary quadratic form $f(x,y,z).$ Mordell points out that these occur in a finite number of parametrizations. The process you mention, stereographic projection around a rational point, does not do a good job of finding primitive integral solutions. Instead, it immediately finds all rational solutions, with no bound on denominators.

The Hessian matrix $H$ of an isotropic ternary form has this feature: there is an integer matrix $P$ and an integer $n$ such that $$ P^T HP = nG \; , $$ where $G$ is the Hessian matrix of $g(x,y,z) = y^2 - zx \; . \;$ Indeed, there are infinitely of these. For a fixed $n,$ there are typically several such $P$ if any.

Let's see, the primitive null vectors of $y^2 - zx$ are precisely $(p^2,pq,q^2).$ Applying $P$ to this (as a column vector) gives a null vector for $H,$ and we get some ability to say when this will be primitive.

I worked this out for isotropic forms of the sort $A(x^2 + y^2 + z^2) - B(yz+zx+xy).$ The number of inequivalent $P$ matrices needed to produce all primitive null vectors can be arbitrarily large. I kept a list somewhere...

The original proof is in Fricke and Klein (1897), where it is mentioned in passing. Different versions have been published over the years. I eventually wrote down a proof using just matrices, gcd and the like.

The twelve matrices $P$ needed for $$ 100(x^2 + y^2 + z^2) -541(yz + zx + xy) =0 $$ This includes something about the order of the (very symmetric) solutions.

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 A = 100       B = 541

    445   1009    430
    430   -149   -134
   -134   -119    445

    478   1003    394
    394   -215   -131
   -131    -47    478

    514    985    349
    349   -287   -122
   -122     43    514

    529    973    328
    328   -317   -116
   -116     85    529

    541    961    310
    310   -341   -110
   -110    121    541

    574    913    253
    253   -407    -86
    -86    235    574

    580    901    241
    241   -419    -80
    -80    259    580

    604    835    184
    184   -467    -47
    -47    373    604

    610    811    166
    166   -479    -35
    -35    409    610

    616    781    145
    145   -491    -20
    -20    451    616

    625    709    100
    100   -509     16
     16    541    625

    628    643     64
     64   -515     49
     49    613    628


   count was  12     end of  A = 100       B = 541
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my proof