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Will Jagy
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Will Jagy
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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.

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

 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
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

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.

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

 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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Will Jagy
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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.