Siegel's Mass Formula for ternary indefinite quadratic forms In his paper "On the theory of indefnite quadratic forms", Siegel gives the formula (Thm. 1)
$$
\mu(S,T)=\prod_p \alpha_p(S,T),
$$
where 


*

*$S$ is an $m\times m$ non singular integral symmetric matrix of signature $(r,m-r)$,

*$T$ is an $n\times n$ integral symmetric matrix, 

*$\mu(S,T)$ is the measure of the representation of $T$ by $S$,

*$\alpha_p(S,T)$ is the $p$-adic density of the representation of $T$ by $S$ ($p$ over the rational primes), 


and 
$$
n\leq r,\qquad n\leq m-r,\qquad 2(n+1) < m.
$$
In my case, $n=1$ thus $T=t$ is just a (non zero) integer number, and $S$ has signature $(m-1,1)$ (or $(1,m-1)$). Hence, the theorem holds for 
$$
m>4.
$$
However, in the fourth page, he added that the formula also holds for $m=4$.
I known that when $m=3$ ($S$ is a ternary quadratic form), the formula doesn't holds in general and strange things happen (see BOROVOI ). 
My question is: Does the formula works for the following particular case?:
$$
S=
\begin{pmatrix}
A & \\ 
  &-a
\end{pmatrix}
$$
where


*

*$A$ is a $2\times2$ integral positive definite symmetric matrix,

*$a$ is a positive integer, and

*$t$ is a negative integer.


I known that this is true for $A=I_2$ and $a=1$ (see this paper).
Thank you in advance.-.
 A: Let me settle the situation with $r^2$ when $r$ is a prime with $r \equiv \pm 3 \pmod 8.$ 
First, suppose we have some representation by the first form,
$$ x^2 - 2 y^2 + 64 z^2 = r^2.   $$ Then
$$  x^2 - 2 y^2 = r^2 - 64 z^2 = (r + 8 z)(r - 8 z).  $$
Now, as $r - 8 z \equiv \pm 3 \pmod 8,$ there is some prime $t \equiv \pm 3 \pmod 8$ with
$$ t^{2 k + 1} \parallel (r - 8 z).   $$ However, we must have
$$  t^{2 m} \parallel (x^2 - 2 y^2),  $$ so
$$ t^{2 l + 1} \parallel (r + 8 z).   $$ As a result, $t | (2 r),$ so $t = r,$ so that $ r | x$ and $r | y.$ We already had $t | (2z),$ so actually $r | z.$ Put these together, $x^2 - 2 y^2 + 64 z^2$ does represent $r^2$ but not primitively. 
This next bit is unexpected, at least in terms of finding an explicit formula. We get a cheap trick out of $$   -9 x^2 + 2 x y + 7 y^2 = (9x + 7 y)(-x+y).  $$ If we take $y=x+1$ the result of the binary is $16 x + 7.$ Add on $2 z^2$ with $z=1$ we actually get $16 x + 9.$
Alright, back to prime $r \equiv \pm 3 \pmod 8.$ It follows that $r^2 \equiv 9 \pmod {16}.$
We have $$ g(x,y,z) = -9x^2 + 2 x y + 7 y^2 + 2 z^2.  $$ Take any number $n \equiv 9 \pmod {16}.$ Then we have an integral, and primitive, representation in
$$ g \left( \frac{n - 9}{16}, \;  \frac{n + 7}{16}, \; 1 \right) = n.   $$
You really do not get this sort of closed form answer with positive forms.
Though you should build a bark of dead men's bones,
And rear a phantom gibbet for a mast,
Stitch creeds together for a sail, with groans
To fill out, blood-stained and aghast;
Although your rudder be a dragon's tail
Long sever'd, yet still hard with agony
Your cordage large uprootings from the skull
Of bald Medusa, certes you would fail
To find the Melancholy - whether she
Dreameth in any isle of Lethe dull. 

