Isotropic ternary forms It is well known that some questions about isotropic ternary forms reduces to the study of the special case $f_0(X)=xz-y^2, X=(x,y,z)$, see page 301 of Cassel's "Rational quadratic forms" (Dover, 2008). In page 303 it is said that given any other general integral isotropic form $f(X)$, there is an integer $m\neq0$ and a nonsingular $3\times 3$ integral matrix $M$ such that 
$$mf(X)=f_0(MX).$$
The question is how to compute explicitly $M$. When $f(X)=ax^2+y^2-z^2$ it is quite easy to find $M$. In general, for $f(X)=ax^2+by^2-z^2$, $a>b>1$, it becomes more difficult. Of course, since the form is isotropic, by Legendre's Theorem, $a$ and $b$ satisfy the three relations:
$$(i) \,\,\, aRb \, , \,\,\,\,\,(ii) \,\,\, bRa \, , \,\,\,\,\,(iii) \,\,\, (-ab)/d^2Rd$$
where $R$ is the equivalence relation $mRn$ if, and only if, $m$ is a square modulo $n$ and $d=d(a,b)=g.c.d(a,b)$. Any general formula to obtain $M$ for a specific isotropic ternary $f(X)$ should take this relations into consideration.
For example, the ternary form $f(X)=19x^2+5y^2-z^2$ is isotropic since $19+45=64$ and $76+5=81$ (this ternary form was considered in Integral orthogonal group for indefinite ternary quadratic form). What is $M$ for this particular ternary form?
 A: The mention of another quadratic form. You can use the standard approach.
In equation $$aX^2+bY^2+cZ^2=qXY+dXZ+tYZ$$    
$a,b,c,q,d,t$ integer coefficients which specify the conditions of the problem. 
For a more compact notation, we introduce a replacement. 
$k=(q+t)^2-4b(a+c-d)$    
$j=(d+t)^2-4c(a+b-q)$   
$n=t(2a-t-d-q)+(2b-q)(2c-d)$     
Then the formula in the general form is:    
$$X=(2n(2c-d-t)+j(q+t-2b\pm\sqrt{k}))p^2+2((d+t-2c)\sqrt{k}\mp{n})ps+$$
$$+(2b-q-t\pm\sqrt{k})s^2$$   
$$Y=(2n(2c-d-t)+j(2(a+c-d)-q-t\pm\sqrt{k}))p^2+2((d+t-2c)\sqrt{k}\mp{ n })ps+$$
$$+(q+t+2(d-a-c)\pm\sqrt{k})s^2$$ 
$$Z=(j(q+t-2b\pm\sqrt{k})-2n(2(a+b-q)-d-t))p^2+$$
$$+2((2(a+b-q)-d-t)\sqrt{k}\mp{n})ps+(2b-q-t\pm\sqrt{k})s^2$$     
And more.    
$$X=(2n(q+t-2b)+k(2c-d-t\pm\sqrt{j}))p^2+2((2b-q-t)\sqrt{j}\mp{n})ps+$$
$$+(d+t-2c\pm\sqrt{j})s^2$$    
$$Y=(2n(2(a+c-d)-q-t)+k(2c-d-t\pm\sqrt{j}))p^2+$$
$$+2((q+t+2(d-a-c))\sqrt{j}\mp{n})ps+(d+t-2c\pm\sqrt{j})s^2$$  
$$Z=(2n(q+t-2b)+k(d+t+2(q-a-b)\pm\sqrt{j}))p^2+2((2b-q-t)\sqrt{j}\mp{n})ps+$$
$$+(2(a+b-q)-d-t\pm\sqrt{j})s^2$$  
$p,s$ are integers and are given us.  
Solution when there is at least one root of a whole. For this it is necessary to consider the possible equivalent forms.  This is done by changing the type $X  \longrightarrow X+kY$ 
The desired coefficients $k$ can be found by solving a simpler equation. As a rule, either the Pell equation or representation of a number as a sum of squares. The equation is much simple than the original.
A: Gauss' Disquisitiones gives a completely algorithmic solution to this question in his development of reduction theory for ternary quadratic forms over the integers (via artful systematic use of his binary reduction theory applied to the xy-part, yz-part, and xz-part in appropriate order, along with those quadratic congruence conditions).  Is there a reason his algorithm is insufficient (other than that it seems nobody has ever "coded it up", for reasons I do not know)?
A: Wednesday morning, June 3.
I will be back home tonight, probably tomorrow I can provide the specific example requested.
The fact in Cassels is correct. It goes back to the seminal book by Fricke and Klein in 1897. It is the first volume of two on automorphic forms. Volume I is about group theory. There is quite a lot. One major thread is the relation between $SL_2 \mathbb Z$ and the integer automorphism group of a form that is already known to be isotropic over $\mathbb Q,$ with integer coefficients. 
Let $H$ be the Hessian matrix of second partial derivatives of your $f_0$ by the chosen variables $x,y,z.$ Let $G$ be the Hessian of $f.$ The matrix $M$ and nonzero integer $m$ are to satisfy
$$ M^t H M = m G. $$
The theorem is that such exist. One constraint is automatic, $m$ must lie in a fixed integer squareclass. If we write $t = \det M,$ $h = \det H,$ $g = \det G,$ we arrive at
$$  h t^2 = m^3 g. $$  If some $t_0$ is the integer of smallest absolute value for which there is an integer solution to $h t_0^2 = m^3 g,$ we know that any successful $t$ must be of the form $t = t_0 w^2.$
There is something special for isotropic ternaries here: if $\gcd(x,y,z) = 1$ and $xz-y^2 = 0$ and either $x \geq 0$ or $z \geq 0,$ then there are integers $u,v,$ not both zero and with $\gcd(u,v) = 1,$
such that
$$ x = u^2; \; \;  y = uv; \; \; z = v^2.  $$
It is for this reason that finding the transition matrix $M$ leads immediately to all rational isotropic vectors of $f,$ just plug in $u,v.$  The hard part, and it is hard, is going from rational solutions to primitive integer solutions. Ummm, I may have lost track of the direction here, the useful direction for what i am describing is $H = r N^T G N,$ where the $H$ refers to $xz-y^2$  and $r$ is rational. One thing this explains completely is why the integer null cone for an isotropic ternary always comes out in terms of binary quadratic forms in some $u,v,$ and the hard part is then "What is the gcd of the three entries $x,y,z?$"
All this, and a good deal more, is in Fricke and Klein in German, in Wilhelm Magnus, Noneuclidean Tessellations and their Groups, and in a conference proceedings edited by Olga Taussky, with major article by Zassenhaus. I cannot remember all details on the Taussky book, I will check later. 
A: Ternary Quadratic Forms and Norms edited by Olga Taussky (1982). Pages 5-30 is William Plesken, Automorphs of Ternary Quadratic Forms. The word automorph is one of the traditional terms for what would now be called a member of the (integer) automorphism group of the form, or rotation group, or orthogonal group.
Since $xz - y^2$ has the mixed coefficient $1,$ we need to double it to get an integral matrix, and this is also the Hessian matrix,
$$
H =
\left(
\begin{array}{rrr}
0 & 0 & 1 \\
0 & -2 & 0 \\
1 & 0 & 0
\end{array}
\right)
$$
This becomes annoying for diagonal forms, where we still double the diagonal entries:
$$
G =
\left(
\begin{array}{rrr}
38 & 0 & 0 \\
0 & 10 & 0 \\
0 & 0 & -2
\end{array}
\right)
$$
In this direction,
take 
$$
N =
\left(
\begin{array}{rrr}
38 & 30 & 16 \\
19 & -25 & -7 \\
-38 & 5 & 9
\end{array}
\right)
$$
for one of infinitely many solutions to
$$ N^T H N = -95 G  $$
give me a few more minutes...
In the other direction, take
$$
M =
\left(
\begin{array}{rrr}
2 & 2 & -2 \\
3 & -8 & -1 \\
11 & -4 & 9
\end{array}
\right)
$$
then
$$ M^T G M = -380 H  $$
A: Not entirely clear where one detail which is not mentioned. Ternary quadratic form always amounts to a Pell equation. For example if you take a fairly simple equation. And set some conditions for the coefficients.
All of numbers can be any character.In Equation:  $qX^2+Y^2=Z^2+a$  
If the ratio is factored so:  $a=(b-c)(b+c)$ 
Then we use the solutions of Pell's equation: $p^2-fs^2=\pm1$ 
where:  $f=(q+1)k^2-2kt-(q-1)t^2$ 
Then the solutions are of the form: 
$$X=2(ck-bt)ps+2(bk^2-(b+c)kt+ct^2)s^2$$ 
$$Y=bp^2+2c(k-t)ps-(b(q-1)k^2+2(b-qc)kt+b(q-1)t^2)s^2$$  
$$Z=cp^2+2b(k-t)ps+(c(q+1)k^2-2(bq+c)kt+c(q+1)t^2)s^2$$ 
All of numbers can be any character.
For the equation: $qX^2+Y^2=Z^2+j$
In the case where a square: $a=\sqrt{\frac{j}{q}}$
Using equation Pell: $p^2-(q+1)s^2=1$
Then the solution can be written:
$$X=2s(s\pm{p})L\pm{ap^2}+2aps\pm{a(q+1)s^2}=bL+af$$
$$Y=(p^2\pm2ps+(1-q)s^2)L\pm{ap^2}+2aps\pm{a(q+1)s^2}=cL+af$$
$$Z=(p^2\pm2ps+(q+1)s^2)L\pm{ap^2}+2a(q+1)ps\pm{a(q+1)s^2}=fL+at$$
$L$ - any integer number given by us.
The most interesting thing is that these numbers are solutions of equations:
$qb^2+c^2=f^2$
$t^2-(q+1)f^2=\pm{q}$
If we use the equation Pell: $p^2-(q+1)s^2=k$
And substituting the solutions in the upper formula, we have solutions of the following equations.
$qb^2+c^2=f^2$
where: $c-b=k$
$t^2-(q+1)f^2=\pm{qk^2}$
True, I use this formula in reverse order. Find solutions of Pell's equation is much more complicated than the simple equations like Pythagorean triples. So find them and then have solutions of Pell's equation. The most interesting thing is that the solution of Pell related to Pythagorean triples.
You can also write a more General formula.  In this case it will be necessary to consider all the possible equivalent forms.  In this case, anyway.  The problem is reduced to solving a Pell equation.  I think it's easier to solve.
This approach makes it quite easy to prove that the curves for triangular numbers is always possible to write the solution.  If the coefficients of the 1-St degree is not equal to 0. And the coefficients of the 2nd degree don't create a trivial situation.  The formula there.  https://math.stackexchange.com/questions/794510/curves-triangular-numbers
