I've enjoyed finding out how to parametrize the integer null vectors of an indefinite ternary with integer coefficients. The observation that an isotropic ternary integrally represents an integer multiple of $y^2 - zx$ goes back to Fricke and Klein (1897 I think).
It is easy to find all ration vectors in the null cone. However, it is fairly difficult to pull out all the primitive integer vectors. If we just multiply through by a denominator (in symbols) from stereographic projection, we typically miss a fixed fraction of solutions.
All told, the outcome is that the integer null vectors are parametrized by a finite number of Pythagorean Triple type formulas, but not necessarily just one.
why not...
This is about ternary quadratic forms with integer coefficients. We use the ordering in http://zakuski.math.utsa.edu/~kap/Lehman_1992.pdf and http://zakuski.math.utsa.edu/~kap/Watson_Representation_1954.pdf I like how Watson writes it, writing $zx$ instead of $xz.$ The ordered sextuple $$ \langle a,b,c, r,s,t \rangle $$ refers to the form $$ f(x,y,z) = a x^2 + b y^2 + c y^2 + r y z + s z x + t x y $$ The Hessian matrix, of second partial derivatives, is
$$ H = \left( \begin{array}{rrr} 2 a & t & s \\ t & 2b & r \\ s & r & 2c \end{array} \right) $$
The discriminant is half the determinant of $H$ and therefore an integer, $$ \Delta = 4abc + rst - a r^2 - b s^2 - c t^2. $$
Oh, right. Given a column vector $$ U = \left( \begin{array}{r} x \\ y \\ z \end{array} \right), $$ we have $$ f(x,y,z) = \frac{1}{2} \; \; U^T H U. $$
We will say that $f$ represents another form $g,$ with Hessian $G,$ if there is an integer matrix $P$ with $\det P \neq 0,$ such that $$ P^T H P = G. $$ Note that $ \det G = \det H \left( \det P \right)^2$ is still nonzero.
Suppose that $f$ is isotropic over the integers. This means that there is some $(u_1, u_2, u_3),$ not all zero, with $f(u_1, u_2, u_3)= 0.$
Theorem: if $f$ is isotropic with nonzero discriminant, then there is a nonzero integer $n$ such that $f$ represents $n (y^2 - zx).$
It took me three steps to prove this, so I will present it that way. The first statement of this that I know is pages 507-508 of
Fricke Klein
This is also on page 303 of Cassels
but is not part of the online preview.
First, we have $$ U = \left( \begin{array}{r} u_1 \\ u_2 \\ u_3 \end{array} \right). $$ The hypothesis is that $$ U^T H U = 0. $$ On the other hand, we are told that the determinant of $H$ is nonzero, which tells us that $HU \neq \vec{0}.$ It follow that there is a nonzero integer vector $$ V = \left( \begin{array}{r} v_1 \\ v_2 \\ v_3 \end{array} \right) $$ such that $$ V^T H U = 0. $$ For example, using the ordinary cross product, we could simply use $V = (HU) \times U.$
We may take any third vector as the final column of a matrix
$$ P = \left( \begin{array}{rrr} u_1 & v_1 & w_1 \\ u_2 & v_2 & w_2 \\ u_3 & v_3 & w_3 \end{array} \right) $$ as long as the determinant is nonzero. The outcome is that, with some integers $i,j,k,l,$ $$ P^T H P = \left( \begin{array}{rrr} 0 & 0 & l \\ 0 & 2i & j \\ l & j & 2k \end{array} \right). $$ That is, $f$ represents $$ \langle 0,i,k,j,l,0 \rangle $$
Next, we take a new matrix $$ Q = \left( \begin{array}{rrr} 1 & 0 & j^2 - 4ik \\ 0 & 1 & -2jl \\ 0 & 0 & 4il \end{array} \right). $$ It took me a long time to find $Q.$ We now have, with some nonzero integers $m,n,$ $$ Q^T P^T H P Q = \left( \begin{array}{rrr} 0 & 0 & n \\ 0 & 2m & 0 \\ n & 0 & 0 \end{array} \right). $$ Finally, we take a diagonal matrix $R,$ $$ R = \left( \begin{array}{rrr} m & 0 & 0 \\ 0 & -n & 0 \\ 0 & 0 & -n \end{array} \right). $$ We reach $$ R^T Q^T P^T H P Q R = \left( \begin{array}{rrr} 0 & 0 & -m n^2 \\ 0 & 2mn^2 & 0 \\ -m n^2 & 0 & 0 \end{array} \right) \; \; = \; \; m n^2 \; \; \left( \begin{array}{rrr} 0 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 0 \end{array} \right) $$
That is, $f$ represents an integer multiple of $y^2 - zx.$ Let us call that $N(y^2 - z x),$ so this $N = m n^2.$
Thursday, 20 October: the matrix entries I came up with for the general case may be much larger than required. On page 28, Cassels asks about $3 x^2 - 2 y^2 - z^2.$ I have a computer program to find these expressions with small numbers. I will not bother doubling the diagonal entries,
% jagy@phobeusjunior:~$ ./homothety_indef 3 -2 -1 0 0 0 0 -24 0 0 24 0 4
$$ \left( \begin{array}{rrr} 2 & 2 & -2 \\ 0 & 2 & 4 \\ 1 & -1 & 1 \end{array} \right) \left( \begin{array}{rrr} 3 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -1 \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & 1 \\ 2 & 2 & -1 \\ -2 & 4 & 1 \end{array} \right) = \left( \begin{array}{rrr} 0 & 0 & 12 \\ 0 & -24 & 0 \\ 12 & 0 & 0 \end{array} \right) $$ That is, $$ \langle 3,-2,-1,0,0,0 \rangle $$ represents $$ \langle 0, -24,0,0,24,0 \rangle $$ or $$ -24 \cdot \langle 0, 1,0,0,-1,0 \rangle. $$
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.
========================================
parisize = 4000000, primelimit = 500509
? x = 120 * u^2 + 229 * u * v + 118 * v^2
%1 = 120*u^2 + 229*v*u + 118*v^2
?
? y = 118 * u^2 + 7 * u * v + 9 * v^2
%2 = 118*u^2 + 7*v*u + 9*v^2
?
? z = 9 * u^2 + 11 * u * v + 120 * v^2
%3 = 9*u^2 + 11*v*u + 120*v^2
? 66 * ( x^2 + y^2 + z^2) - 115 * ( y * z + z * x + x * y)
%4 = 0
?
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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
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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