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 $N = m n^2.$


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
$$ -24 \cdot \langle 0, 1,0,0,-1,0 \rangle. $$


  [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