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This shows the six Pythagorean Triple type parametrizations needed for $$ 66 (x^2 + y^2 + z^2) - 115(yz + zx + xy) =0 $$ The first matrix means $$ x = 120 u^2 + 229 uv + 118 v^2; \; \; y = 118 u^2 + 7 u v + 9 v^2; \; \; z = 9 u^2 + 11 u v + 120 v^2. $$ This is up to order (lots of symmetry) and negating all three $xyz.$ $u,v$ are not required of the same $\pm$ sign. Indeed, we rename and negate so that $x \geq |y| \geq |z|.$ Otherwise many more recipes (such matrices) might be needed.

This shows the six Pythagorean Triple type parametrizations needed for $$ 66 (x^2 + y^2 + z^2) - 115(yz + zx + xy) =0 $$ The first matrix means $$ x = 120 u^2 + 229 uv + 118 v^2; \; \; y = 118 u^2 + 7 u v + 9 v^2; \; \; z = 9 u^2 + 11 u v + 120 v^2. $$ This is up to order (lots of symmetry) and negating all three $xyz.$ $u,v$ are not required of the same $\pm$ sign. Indeed, we rename and negate so that $x \geq |y| \geq |z|.$ Otherwise many more recipes (such matrices) might be needed. Oh, there is something very unusual caused by the extreme symmetry for this problem: in all cases, we may demand $u,v \geq 0.$ Rare behavior. The other oddity, that we always get $x,y,z \geq 0,$ is simply that $2 \cdot 66 > 115,$ and says only that the null cone happens to lie entirely inside the positive octant (and the negative octant).

jagy@phobeusjunior:~$ ./isotropy 66 115

 A = 66       B = 115

    120    229    118
    118      7      9
      9     11    120

    129    227    108
    108    -11     10
     10     31    129

    145    211     84
     84    -43     18
     18     79    145

    150    199     73
     73    -53     24
     24    101    150

    153    187     64
     64    -59     30
     30    119    153

    154    181     60
     60    -61     33
     33    127    154


  end of  A = 66       B = 115

 B - 2 A =  -17       B - A = 49      B + 2 A =  247

  gcd( 4B-4A, B+2A) =  1

 lambda =  247  t = 1  lambda t  = 247
 2 alpha - beta + 2 gamma = 247
 alpha^2 + (alpha - beta + gamma)^2  + gamma^2 = 28405
    beta^2 - 4 alpha gamma = -4199
 matrix determinants  = +/-  2989441 = 7^2 * 13^2 * 19^2
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jagy@phobeusjunior:~$ ./isotropy 66 115

 A = 66       B = 115

    120    229    118
    118      7      9
      9     11    120

    129    227    108
    108    -11     10
     10     31    129

    145    211     84
     84    -43     18
     18     79    145

    150    199     73
     73    -53     24
     24    101    150

    153    187     64
     64    -59     30
     30    119    153

    154    181     60
     60    -61     33
     33    127    154


  end of  A = 66       B = 115 


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

===================================================

    120    118      9      < 120, 229, 118 >      1  0    
    129    108     10      < 129, 227, 108 >      1  0    
    145     84     18      < 145, 211, 84 >      1  0    
    150     73     24      < 150, 199, 73 >      1  0    
    153     64     30      < 153, 187, 64 >      1  0    
    154     60     33      < 154, 181, 60 >      1  0    
    395    314     32      < 154, 181, 60 >      1  1    
    404    302     35      < 153, 187, 64 >      1  1    
    422    275     44      < 150, 199, 73 >      1  1    
    440    242     59      < 145, 211, 84 >      1  1    
    464    170    107      < 129, 227, 108 >      1  1    
    467    140    134      < 120, 229, 118 >      1  1    
    880    783     66      < 153, 187, 64 >      1  2     
   1038    540    151      < 154, 181, 60 >      2  1    
   1056    495    178      < 120, 229, 118 >      2  1    
   1086    375    268      < 145, 211, 84 >      2  1    
   1764   1290    157      < 153, 187, 64 >      1  3    
   1782   1264    165      < 129, 227, 108 >      1  3    
   1800   1237    174      < 154, 181, 60 >      1  3    
   1869   1122    220      < 120, 229, 118 >      1  3    
   2020    669    522      < 150, 199, 73 >      3  1    
   2022    645    544      < 145, 211, 84 >      3  1    
   2310   2211    172      < 153, 187, 64 >      2  3    
   2602   1851    240      < 145, 211, 84 >      2  3    
   2654   2333    200      < 145, 211, 84 >      1  4    
   2712   1675    306      < 154, 181, 60 >      3  2    
   2836   1422    435      < 150, 199, 73 >      3  2    
   2916   1182    595      < 120, 229, 118 >      2  3    
   2924   1973    290      < 120, 229, 118 >      1  4    
   2955    946    792      < 129, 227, 108 >      3  2    
   3005   1838    344      < 154, 181, 60 >      1  4    
   3260   1109    818      < 153, 187, 64 >      4  1 

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This shows the six Pythagorean Triple type parametrizations needed for $$ 66 (x^2 + y^2 + z^2) - 115(yz + zx + xy) =0 $$ The first matrix means $$ x = 120 u^2 + 229 uv + 118 v^2; \; \; y = 118 u^2 + 7 u v + 9 v^2; \; \; z = 9 u^2 + 11 u v + 120 v^2. $$ This is up to order (lots of symmetry) and negating all three $xyz.$ $u,v$ are not required of the same $\pm$ sign.

This shows the six Pythagorean Triple type parametrizations needed for $$ 66 (x^2 + y^2 + z^2) - 115(yz + zx + xy) =0 $$ The first matrix means $$ x = 120 u^2 + 229 uv + 118 v^2; \; \; y = 118 u^2 + 7 u v + 9 v^2; \; \; z = 9 u^2 + 11 u v + 120 v^2. $$ This is up to order (lots of symmetry) and negating all three $xyz.$ $u,v$ are not required of the same $\pm$ sign. Indeed, we rename and negate so that $x \geq |y| \geq |z|.$ Otherwise many more recipes (such matrices) might be needed.