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][1]
This is also on page 303 of [Cassels][2]
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 $N = m n^2.$
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. 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).
========================================
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
?
=========================================
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
========================================================
[1]: https://books.google.com/books?id=H5kLAAAAYAAJ&printsec=frontcover&dq=Vorlesungen%20%C3%BCber%20die%20Theorie%20der%20automorphen%20%20Functionen&hl=en&sa=X&ved=0CCAQ6AEwAGoVChMIscb4rNvYxgIVSCiICh3qMwQX#v=onepage&q=muesli&f=false
[2]: http://store.doverpublications.com/0486466701.html