5
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Preface: the most natural way to take one isotropic vector for an indefinite quadratic form and find others is to use stereographic projection. This gives a parametrization in the same $n$ variables as the quadratic form. After clearing denominators, there is also not much control over the gcd of the resulting integers. So, although finding an integer multiple of every primitive solution is guaranteed, we may not be entirely sure we have found all primitive solutions with entries up to some bound in absolute value.

There is a trick for indefinite ternary forms, which leads to a parametrization by two parameters, with considerable control of the gcd's.

Most of this appears in my answers to Isotropic ternary forms

Question: is it true that the primitive integer solutions to $$ A(x^2 + y^2 + z^2) - B (yz+zx+xy) =0 $$ can all be parametrized by a finite number of solutions as below, in shorthand $R_j U?$ The calculations are awfully convincing, but I have proved only a few. In case anyone gets interested, i wrote out the proof for $A=2, B=113,$ about twenty four pages pdf.

Once we have integers $B > A > 0,$ a necessary and sufficient condition that the form be isotropic in $\mathbb Q,$ and therefore $\mathbb Z,$ is that both $B-A$ and $B+2A$ have integer expressions as $s^2 + 3 t^2.$

There is an interesting alternative, method goes back to Fricke and Klein, gives a two variable parametrization, and can be adjusted to deal with GCD's. There is a complete answer to this, finding all primitive solutions, meaning $\gcd(x,y,z) = 1.$

We begin with finding all primitive solutions to $y^2 - z x = 0.$ If $g = \gcd(x,z) > 1,$ then $g^2 | y^2$ and $g | y,$ so $g | \gcd(x,y,z).$ However, $\gcd(x,y,z) = 1.$ So, $\gcd(x,z) = 1.$ Since $xz = y^2,$ either $x=u^2, z=v^2,$ or $x=-u^2,z=-v^2,$ in either case with $\gcd(u,v) = 1.$ That is, possibly by changing from $(x,y,z)$ to $(-x,-y,-z)$ so as to arrange $x \geq 0,$ all primitive solutions are $$ x = u^2, y = u v, z = v^2. $$

Next, the quadratic form is $X^T G X / 2,$ where $$ G = \left( \begin{array}{rrr} 2a & d & e \\ d & 2b & f \\ e & f &2c \end{array} \right) $$ and $$ X = \left( \begin{array}{r} x \\ y \\ z \end{array} \right) $$

The quadratic form $y^2 - z x$ is $X^T H X / 2,$ where $$ H = \left( \begin{array}{rrr} 0 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 0 \end{array} \right) $$ It is a theorem in Fricke and Klein (1897), pages 507-508, that, because the quadratic form has at least one integer solution, in which $(x,y,z)$ are not all zero, there exists a square matrix of integers $R$ and a nonzero integer $n$ such that $$ R^T G R = n H. $$ As $R$ has an inverse, we can take $S$ to have integral entries and minimize positive $k$ in $$ RS = SR = k I. $$

We already know that we can take all solutions of $y^2 - z x = 0$ as the column vector $$ U = \left( \begin{array}{r} u^2 \\ uv \\ v^2 \end{array} \right) $$ for relatively prime $(u,v).$ That is, $U^T H U = 0,$ and all solutions are a scalar multiple of $U.$

What happens if $X^T G X = 0,$ the "solutions" we want, with gcd one? Well, $R^T G R = n H,$ so $S^T R^T G R S = n S^T H S,$ so $$ G = \frac{n}{k^2} S^T H S, $$ and $X^T G X = 0$ says $$ X^T S^T H S X = 0. $$ We have already shown that there is some integer $w$ with $$ SX = w U. $$ This gives us $RSX = w RU$ and $kX = w RU.$ Now, as $\gcd(x,y,z) = 1,$ there is a row vector $A = (\alpha,\beta,\gamma)$ with $AX = 1.$ This tells us $k = w ARU.$ As $ARU$ is some integer, $w | k,$ and the earlier $kX = w RU$ becomes $$ X = \frac{1}{h} R U $$ for $h = \frac{k}{w} \in \mathbb Z.$ Furthermore, as $ RS = SR = k I, $ we know $k | \det R,$ so $h | k$ tells us $h | \det R.$ One may leave it this way: list the divisors of $\det R,$ including $\det R$ itself. For each primitive pair $(u,v),$ produce the column vector $RU,$ which will be a solution but perhaps not primitive. Divide out by the gcd of the entries of $RU.$ All integer primitive solutions are given by $$ X = RU/ g_1, $$ where $g_1$ is the gcd of the three entries of $RU.$ It is worth emphasizing that $g_1$ is a divisor of $\det R.$ Also, we get some explicit bounds, as $$ |X|^2 = \frac{1}{g_1^2} U^T R^T R U, $$ since $R$ is nonsingular integer and $R^T R$ is symmetric positive definite. So, no matter what, we have a way to find all primitive solutions $X$ with some $|X| \leq \mbox{bound}$ by taking $|u|, |v|$ up to some other bound we can figure out. $$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$ A more interesting alternative: for each divisor of $\det R,$ we may rewrite the eventual primitive solution with that gcd as a new recipe, $R_1 U$ for a new integer matrix $R_1$ that also solves $R_1^T G R_1 = n H.$

The example I like to show is solving $$ 2(x^2 + y^2 + z^2) - 113(yz + zx + xy)=0, $$ four "recipes," $$ \left( \begin{array}{r} x \\ y \\ z \end{array} \right) = \left( \begin{array}{r} 37 u^2 + 51 uv + 8 v^2 \\ 8 u^2 -35 uv -6 v^2 \\ -6 u^2 + 23 uv + 37 v^2 \end{array} \right) $$

$$ \left( \begin{array}{r} x \\ y \\ z \end{array} \right) = \left( \begin{array}{r} 32 u^2 + 61 uv + 18 v^2 \\ 18 u^2 -25 uv -11 v^2 \\ -11 u^2 + 3 uv + 32 v^2 \end{array} \right) $$

$$ \left( \begin{array}{r} x \\ y \\ z \end{array} \right) = \left( \begin{array}{r} 38 u^2 + 45 uv + 4 v^2 \\ 4 u^2 -37 uv -3 v^2 \\ -3 u^2 + 31 uv + 38 v^2 \end{array} \right) $$

$$ \left( \begin{array}{r} x \\ y \\ z \end{array} \right) = \left( \begin{array}{r} 29 u^2 + 63 uv + 22 v^2 \\ 22 u^2 -19 uv -12 v^2 \\ -12 u^2 -5 uv + 29 v^2 \end{array} \right) $$

For all four recipes, $$ x^2 + y^2 + z^2 = 1469 \left( u^2 + uv + v^2 \right)^2 $$ Since $u^2 + uv + v^2 \geq 3 u^2 / 4$ and $u^2 + uv + v^2 \geq 3 v^2 / 4,$ this gives us explicit bounds on the absolute values of $u,v$ that gives us all (primitive) solutions of $ 2(x^2 + y^2 + z^2) = 113 (yz+zx+xy) $ with the absolute values of $x,y,z$ up to a desired bound.

Indeed, we were able to choose all four coefficient matrices with this pattern: $$ R = \left( \begin{array}{ccc} \alpha & \beta & \gamma \\ \gamma & - \beta + 2 \gamma & \alpha - \beta + \gamma \\ \alpha - \beta + \gamma & 2 \alpha - \beta & \alpha \end{array} \right) $$

The rows constitute a cycle of three neighboring, but not reduced, binary quadratic forms under the action of the matrix $$ P = \left( \begin{array}{rr} 0 & 1 \\ -1 & -1 \end{array} \right), $$ where $P^3 = I.$ As soon as we write $X = RU$ we get the identity $$ x^2 + y^2 + z^2 = \left( \alpha^2 + (\alpha - \beta + \gamma)^2 + \gamma^2 \right) \cdot \left( u^2 + uv + v^2 \right)^2 $$

In all four cases we simply discard occurrences when the resulting $x,y,z$ have a common factor. With the understanding that we negate all $x,y,z$ so that the entry with largest absolute value is positive, then sort so that $$ x \geq |y| \geq |z|, $$ here are the answers with maximum up to $1200$

jagy@phobeusjunior:~$ ./isotropy_binaries_combined 2 113 1200 | sort -n
             x      y     z          first line       u   v
             29     22    -12      < 29, 63, 22 >      1  0    
             32     18    -11      < 32, 61, 18 >      1  0    
             37      8     -6      < 37, 51, 8 >      1  0    
             38      4     -3      < 38, 45, 4 >      1  0    
            188    171    -86      < 37, 51, 8 >      1  2    
            211    144    -82      < 38, 45, 4 >      1  2    
            226    123    -76      < 32, 61, 18 >      1  2    
            243     94    -64      < 29, 63, 22 >      1  2    
            246     88    -61      < 38, 45, 4 >      2  1    
            258     59    -44      < 37, 51, 8 >      2  1    
            264     38    -29      < 29, 63, 22 >      2  1    
            268     11     -6      < 32, 61, 18 >      2  1    
            396    262   -151      < 37, 51, 8 >      1  3    
            432    209   -134      < 38, 45, 4 >      1  3    
            472    129    -94      < 29, 63, 22 >      3  1    
            489     76    -58      < 32, 61, 18 >      3  1    
            516    458   -233      < 38, 45, 4 >      2  3    
            526    447   -232      < 37, 51, 8 >      2  3    
            628    311   -198      < 38, 45, 4 >      3  2    
            656    262   -177      < 32, 61, 18 >      2  3    
            671    232   -162      < 37, 51, 8 >      3  2    
            692    183   -134      < 29, 63, 22 >      2  3    
            726     47    -32      < 32, 61, 18 >      3  2    
            727     36    -22      < 29, 63, 22 >      3  2    
            804    787   -382      < 32, 61, 18 >      1  5    
            894    688   -373      < 29, 63, 22 >      1  5    
            953    946   -456      < 38, 45, 4 >      3  4    
           1034    492   -317      < 37, 51, 8 >      1  5    
           1062    443   -296      < 29, 63, 22 >      5  1    
           1102    363   -256      < 38, 45, 4 >      1  5    
           1123    314   -228      < 32, 61, 18 >      5  1    
           1159   1046   -528      < 32, 61, 18 >      1  6    
           1179    118    -88      < 38, 45, 4 >      5  1    
           1188     19      2      < 37, 51, 8 >      5  1    
           1199   1002   -524      < 29, 63, 22 >      1  6
             x      y     z          first line       u   v

I should probably point out that, while it was quite easy (after writing the C++ programs) to make a list of primitive solutions to $ 2(x^2 + y^2 + z^2) - 113(yz + zx + xy)=0 $ with $x \geq |y| \geq |z|$ for $x \leq 1200,$ and just as easy to identify the four square matrices $R_1,R_2,R_3,R_4$ used above, it was quite a big job to prove that these really do give all (ordered) primitive solutions. I have a pdf of the whole business in detail, about twenty pages Latex. Oh: in the above, we may always take $u,v \geq 0.$ It is a reasonable conjecture that the problem $ A(x^2 + y^2 + z^2) - B(yz + zx + xy)=0, $ with $\gcd(A,B)=0,$ $B > A > 0,$ and both $B-A$ and $B + 2A$ expressible in integers as $s^2 + 3 t^2,$ always works out with a finite number of such $R_i.$ No proof.

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  • 1
    $\begingroup$ What's the point in a permanent selection of numbers? $A=2$ ; $B=113$ You set their value via the parameters $s,t$ . And now using them as parameters to write the parametrization of solutions. $\endgroup$ – individ Dec 11 '15 at 17:30
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Another example, with three of the "recipes" required. All these problems I have checked needed either $2^k$ or $3 \cdot 2^k$ such $R$ matrices.

jagy@phobeusjunior:~$ ./isotropy 1 50

 A = 1       B = 50

     19     42     15
     15    -12     -8
     -8     -4     19

     24     36      7
      7    -22     -5
     -5     12     24

     25     32      4
      4    -24     -3
     -3     18     25


  end of  A = 1       B = 50

 B - 2 A =  48       B - A = 49      B + 2 A =  52

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

 lambda =  13  t = 2  lambda t  = 26
 2 alpha - beta + 2 gamma = 26
 alpha^2 + (alpha - beta + gamma)^2  + gamma^2 = 650
    beta^2 - 4 alpha gamma = 624
 matrix determinants  = +/-  16562 = 2 * 7^2 * 13^2
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= 

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
jagy@phobeusjunior:~$ ./isotropy_just_ordered 1 50 1000
   19   15   -8
   24    7   -5
   25    4   -3
   61   40  -23
   67   31  -20
   76    7   -5
  124  115  -57
  163   60  -41
  276  157  -95
  280  151  -93
  328  321 -155
  331   15   -8
  375  268 -149
  433  181 -120
  460  123  -89
  463  115  -84
  483   31  -20
  487  184 -125
  535   19   -8
  604  447 -245
  655  379 -228
  673  589 -300
  700  307 -201
  700  631 -317
  712  285 -191
  760  483 -281
  787   60  -41
  789   40  -23
  811  652 -345
  820  393 -251
  879  280 -197
  915  184 -137
  931  492 -305
  943   24   -5
  951  460 -293
  955  223 -164
  961  205 -152
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

jagy@phobeusjunior:~$ ./isotropy_binaries_combined 1 50 1000 | sort -n
                 19     15     -8      < 19, 42, 15 >      1  0    
                 24      7     -5      < 24, 36, 7 >      1  0    
                 25      4     -3      < 25, 32, 4 >      1  0    
                 61     40    -23      < 25, 32, 4 >      1  1    
                 67     31    -20      < 24, 36, 7 >      1  1    
                 76      7     -5      < 19, 42, 15 >      1  1    
                124    115    -57      < 24, 36, 7 >      1  2    
                163     60    -41      < 19, 42, 15 >      1  2    
                276    157    -95      < 25, 32, 4 >      1  3    
                280    151    -93      < 19, 42, 15 >      1  3    
                328    321   -155      < 25, 32, 4 >      2  3    
                331     15     -8      < 24, 36, 7 >      3  1    
                375    268   -149      < 24, 36, 7 >      2  3    
                433    181   -120      < 25, 32, 4 >      3  2    
                460    123    -89      < 24, 36, 7 >      3  2    
                463    115    -84      < 19, 42, 15 >      2  3    
                483     31    -20      < 19, 42, 15 >      3  2    
                487    184   -125      < 19, 42, 15 >      4  1    
                535     19     -8      < 24, 36, 7 >      4  1    
                604    447   -245      < 19, 42, 15 >      1  5    
                655    379   -228      < 24, 36, 7 >      1  5    
                673    589   -300      < 25, 32, 4 >      3  4    
                700    307   -201      < 19, 42, 15 >      5  1    
                700    631   -317      < 24, 36, 7 >      2  5    
                712    285   -191      < 25, 32, 4 >      1  5    
                760    483   -281      < 24, 36, 7 >      3  4    
                787     60    -41      < 24, 36, 7 >      5  1    
                789     40    -23      < 25, 32, 4 >      5  1    
                811    652   -345      < 19, 42, 15 >      1  6    
                820    393   -251      < 25, 32, 4 >      4  3    
                879    280   -197      < 24, 36, 7 >      4  3    
                915    184   -137      < 19, 42, 15 >      3  4    
                931    492   -305      < 24, 36, 7 >      1  6    
                943     24     -5      < 19, 42, 15 >      4  3    
                951    460   -293      < 19, 42, 15 >      6  1    
                955    223   -164      < 19, 42, 15 >      5  2    
                961    205   -152      < 25, 32, 4 >      5  2    
    jagy@phobeusjunior:~$
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Here's another one I really did prove, $x^2 + y^2 + z^2 - 5(yz+zx+xy)=0.$ This one requires just one recipe, $$ \left( \begin{array}{c} 5 u^2 + 9 uv + 3 v^2 \\ 3 u^2 - 3 u v - v^2 \\ - u^2 + uv + 5 v^2 \end{array} \right) $$ with the understanding that these are permuted, and possibly all mulitplied by $-1,$ to arrange $$ x \geq |y| \geq |z|. $$ As with the other problems, we just discard any triples where $\gcd(x,y,z) \neq 1.$ That is the whole thing, really, with just one "recipe" we are guaranteed an integer multiple of every primitive integer solution; the question is how to get the primitive ones themselves.

jagy@phobeusjunior:~$ ./isotropy 1 5

 A = 1       B = 5

      5      9      3
      3     -3     -1
     -1      1      5


  end of  A = 1       B = 5

 B - 2 A =  3       B - A = 4      B + 2 A =  7

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

 lambda =  7  t = 1  lambda t  = 7
 2 alpha - beta + 2 gamma = 7
 alpha^2 + (alpha - beta + gamma)^2  + gamma^2 = 35
    beta^2 - 4 alpha gamma = 21
 matrix determinants  = +/-  196 = 2^2 * 7^2
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
jagy@phobeusjunior:~$ 

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
jagy@phobeusjunior:~$ 
    jagy@phobeusjunior:~$ ./isotropy_just_ordered 1 5 500
    5    3   -1
   17    5   -1
   41    5    3
   59   47  -15
   75   17   -1
   89   83  -25
  101   47  -15
  111   17    5
  129  125  -37
  173   59  -15
  185  131  -43
  185  167  -51
  201   83  -25
  215   41    3
  227   41    5
  237   89  -25
  251  215  -67
  255  131  -43
  293  255  -79
  311  125  -37
  327  269  -85
  335  129  -37
  353   75   -1
  381  257  -85
  383  101  -15
  395  167  -51
  425  419 -123
  453  335 -109
  461   75   17
  479  257  -85
  489  215  -67
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

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

jagy@phobeusjunior:~$ ./isotropy_binaries_combined 1 5 500 | sort -n
                  5      3     -1      < 5, 9, 3 >      1  0    
                 17      5     -1      < 5, 9, 3 >      1  1    
                 41      5      3      < 5, 9, 3 >      2  1    
                 59     47    -15      < 5, 9, 3 >      1  3    
                 75     17     -1      < 5, 9, 3 >      3  1    
                 89     83    -25      < 5, 9, 3 >      1  4    
                101     47    -15      < 5, 9, 3 >      2  3    
                111     17      5      < 5, 9, 3 >      3  2    
                129    125    -37      < 5, 9, 3 >      1  5    
                173     59    -15      < 5, 9, 3 >      5  1    
                185    131    -43      < 5, 9, 3 >      2  5    
                185    167    -51      < 5, 9, 3 >      1  6    
                201     83    -25      < 5, 9, 3 >      3  4    
                215     41      3      < 5, 9, 3 >      4  3    
                227     41      5      < 5, 9, 3 >      5  2    
                237     89    -25      < 5, 9, 3 >      6  1    
                251    215    -67      < 5, 9, 3 >      1  7    
                255    131    -43      < 5, 9, 3 >      3  5    
                293    255    -79      < 5, 9, 3 >      2  7    
                311    125    -37      < 5, 9, 3 >      7  1    
                327    269    -85      < 5, 9, 3 >      1  8    
                335    129    -37      < 5, 9, 3 >      4  5    
                353     75     -1      < 5, 9, 3 >      5  4    
                381    257    -85      < 5, 9, 3 >      3  7    
                383    101    -15      < 5, 9, 3 >      7  2    
                395    167    -51      < 5, 9, 3 >      8  1    
                425    419   -123      < 5, 9, 3 >      2  9    
                453    335   -109      < 5, 9, 3 >      3  8    
                461     75     17      < 5, 9, 3 >      7  3    
                479    257    -85      < 5, 9, 3 >      4  7    
                489    215    -67      < 5, 9, 3 >      9  1    
    jagy@phobeusjunior:~$
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The same formal record. If we look for a parameterization not in 2 and in 3 option the problem can be solved quite simply.

For the equation.

$$aX^2+bY^2+cZ^2=dXY+eXZ+fYZ$$

If you know any solution $(x,y,z)$ of this equation. Then the formula for the solution of the equation can be written immediately.

$$X=(dy+ez-ax)p^2+(fz-2by)ps+bxs^2+cxt^2+(fy-2cz)pt-fxst$$

$$Y=ayp^2+(ez-2ax)ps+(dx+fz-by)s^2+cyt^2-eypt+(ex-2cz)st$$

$$Z=azp^2-dzps+bzs^2+(ex+fy-cz)t^2+(dy-2ax)pt+(dx-2by)st$$

$p,s,t - $ any integers.

It is seen that such formulas can be written infinitely many. If so like to use the well-known decision, it is better to write like this. This will allow to solve the equation.

$$A(x^2+y^2+z^2)=B(xy+xz+yz)$$

It is easy to see that there are solutions for any $A$. Ask yourself the number $c$. And place into factors.

$$c^2+A=ab$$

Then the coefficient is set to $B$ us so.

$$B=a^2+b^2-2(a+b)c+3c^2$$

Finding all the factors of $c,a,b$ you can write a formula for the parameterization of the solution of this equation.

$$x=c(ab-c^2)(k^2+s^2)+((a+b)(a^2+b^2)-(4a^2+5ab+4b^2)c+7(a+b)c^2-$$

$$-5c^3)p^2-(a^2+b^2-2(a+b)c+3c^2)cks+(b(b^2-a^2)-c(a^2+3b^2)+$$

$$+(4a+5b)c^2-5c^3)pk+(a(a^2-b^2)-c(b^2+3a^2)+(5a+4b)c^2-5c^3)ps$$

$$y=(a-c)(ab-c^2)(p^2+s^2)+(b^3-(a+2b)bc+(a+3b)c^2-c^3)k^2-$$

$$-(a-c)(a^2+b^2-2(a+b)c+3c^2)ps+(-2ab^2+c(a+b)^2-2ac^2+c^3)ks+$$

$$+(b(b^2+a^2)-(a^2+4ab+3b^2)c+(2a+5b)c^2-c^3)kp$$

$$z=(b-c)(ab-c^2)(p^2+k^2)+(a^3-(b+2a)ac+(3a+b)c^2-c^3)s^2-$$

$$-(b-c)(a^2+b^2-2(a+b)c+3c^2)pk+(a(a^2+b^2)-(3a^2+4ab+b^2)c+$$

$$+(5a+2b)c^2-c^3)ps+(-2ba^2+c(a+b)^2-2bc^2+c^3)ks$$

The transition to 3 parameters $p,s,k - $ Allows not to use too much, but to write formulas.

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