On the multiplicativity of integers of the form $2x^2+51xy-32y^2$?

Question

Given the quadratic form,

$$F(x,y)=2x^2+51xy-32y^2$$

this has positive discriminant $d = 2857$ and class number 3 (A094612). It can be shown that integers of this form $F(x,y)$ are closed under multiplication,

$$(2p^2+51pq-32q^2)(2r^2+51rs-32s^2)=2x^2+51xy-32y^2$$

I have an elementary proof (involving much larger coefficients than Brahmagupta's identity), but what would be the traditional number theoretic way to show that it is indeed the case, and can be extrapolated for all class number 3 discriminants?

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Edit:

In response to Jagy's answer, here is the polynomial identity,

$$(2p^2+51pq-32q^2)(2r^2+51rs-32s^2)=2x^2+51xy-32y^2$$

where,

$$x = p(127404r + 3326866s) + q(3326866r + 86873547s)\\ y= p(-4879r - 127404s) - q(127404r + 3326866s)$$

which one can easily verify with Mathematica and others.

Answer 1

The chapter from Buell showing these calculations:

http://zakuski.math.utsa.edu/~jagy/indefinite_binary_Buell.pdf

Dickson (1929) Introduction tp the Theory of Numbers, gives some helpful things, especially Theorem 85 on page 111, due to Lagrange, that all small (in absolute value) numbers that are (primitively) represented by an indefinite form, really do occur as coefficients in its chain of reduced forms

I guess, for class number this small, all you need is to tell when two indefinite forms are distinct. Representation of a single prime tells us that sort of thing.

Buell, section 4.4, pages 74-75. If $p$ is an odd prime that does not divide the discriminant, if $p$ is represented by some form of the discriminant, the only other form that represents $p$ is the "opposite" form. For odd discriminant $1 \pmod 8$ the prime $2$ is represented by one pair of opposite forms. Note $x^2 + y^2$ is a special case here. Otherwise, for even discriminant, the prime $2$ is represented by a single "ambiguous" form, that meaning it is equivalent to its opposite.

Write them with 53 in the middle, then use Dirichlet composition

$$ \langle 1, 53, -12 \rangle \; \; , \; \; \langle 2, 53, -6 \rangle \; \; , \; \; \langle 4, 53, -3 \rangle \; \; , \; \; $$

Then it is a separate problem to find the mapping showing (4,53, -3) is equivalent to ( 2, -53, -6 ) or (-6, 53, 2) Give me a few minutes

First, (4,53,-3) really does represent 2

0  form   4 53 -3   delta  -17
1  form   -3 49 38   delta  1
2  form   38 27 -14   delta  -2
3  form   -14 29 36   delta  1
4  form   36 43 -7   delta  -6
5  form   -7 41 42   delta  1
6  form   42 43 -6   delta  -8
7  form   -6 53 2   delta  26
8  form   2 51 -32   delta  -1

Added: the smallest coefficients are going to come from line 7, form(-6,53, 2). Meanwhile, once you know that (4,53,-3) does not represent 1, it is either (2,53,-6) or its opposite, so that finishes your multiplication question,,,

and really does not represent 1

represented numbers of small absolute value 
     -42     -38     -36     -34     -32     -31     -28     -26     -24     -21
     -18     -17     -16     -14     -13     -12      -9      -7      -6      -4
      -3      -2       2       3       4       6       7       9      12      13
      14      16      17      18      21      24      26      28      31      32
      34      36      38      42

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

Here is how you get   (-6, 53, 2)   out of (4,53, -3)   

with coefficients of modest size

? h = [ 8,53; 53,-6]
%1 = 
[ 8 53]
[53 -6]
? d = matdet(h)
%2 = -2857
?
? r = [ -31, 275; -550, 4879]
%3 = 
[ -31  275]
[-550 4879]
? 
? 
? 
? rt = mattranspose(r)
%4 = 
[-31 -550]
[275 4879]
?
? h
%5 = 
[ 8 53]
[53 -6]
? rt * h * r
%6 = 
[-12 53]
[ 53  4]
? 
   the   form for this Hessian is -6, 53, 2

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$