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. There is a trick for indefinite ternary forms, which leads to a parametrization by two parameters.
Most of this appears in my answers to http://mathoverflow.net/questions/208158/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,"
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
37 u^2 + 51 uv + 8 v^2 \\
8 u^2 -35 uv -6 v^2 \\
-6 u^2 + 23 uv + 37 v^2
\end{array}
\right)
$$
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
32 u^2 + 61 uv + 18 v^2 \\
18 u^2 -25 uv -11 v^2 \\
-11 u^2 + 3 uv + 32 v^2
\end{array}
\right)
$$
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
38 u^2 + 45 uv + 4 v^2 \\
4 u^2 -37 uv -3 v^2 \\
-3 u^2 + 31 uv + 38 v^2
\end{array}
\right)
$$
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
29 u^2 + 63 uv + 22 v^2 \\
22 u^2 -19 uv -12 v^2 \\
-12 u^2 -5 uv + 29 v^2
\end{array}
\right)
$$
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.