A conjecture on products/composition of Pell forms Based on a few brute-force calculations, I've formulated the following.
Conjecture. Let $x,y,u,v,p,q,a,b,c \ge 2$ be integers such that
$$
(x^2+ay^2)(u^2+bv^2) = p^2+cq^2,
$$
and write
\begin{align}
 g_1 &= \gcd(x,y)\gcd(u,v)\gcd(p,q),  \\
 g_2 &= \gcd(a,xy)\gcd(b,uv)\gcd(c,pq),  \\
 g_3 &= \gcd(a,b)\gcd(a,c)\gcd(b,c).
\end{align}
Then $g_1g_2g_3>1$.
Is the conjecture true? What would be a good way of trying to prove it? Alternatively, what would be a good way of narrowing the search for counterexamples (e.g., modular considerations, etc.)?
Note that it isn't enough to simply restrict $g_1 = 1$, since
$$(x,a,y,u,b,v,p,c,q)=(2, 2, 3, 3, 7, 2, 2, 10, 9)$$
is a solution with $g_1=1$ (but $g_2g_3>1$).

ORIGINAL POST:
I have an equation of the form
$$
(x^2+ay^2)(u^2+bv^2) = p^2+cq^2,
$$
where $a,b,c$ are pairwise relatively prime squarefree integers.
Are there any important results regarding this special equation?
EDIT: In Composition of Binary Forms and the Foundation of Mathematics, Harold M. Edwards says that Gauss [ca. Art 235/236 of Disquisitiones Arithmeticae] proved the following theorem:
If $f$ and $\phi$ can be composed, the ratio of their determinants must be a ratio of squares.
Is this really a necessary and sufficient condition? If so, doesn’t the fact that my equation
$$
  (x^2+ay^2)(u^2+bv^2) = p^2+cq^2
$$
have two forms on the left-hand side being composed into the single form on the right-hand side then demand that the determinants of the two left-hand forms must be a ratio of squares?
EDIT: Cross-posted to MO a more specific question about the existence or derivation of a complete solution.
 A: This is an attempt to show that
$$(x^2+ay^2)(u^2+bv^2)=p^2+cq^2 \tag{1}$$ 
has solutions, without conditions on $a,b,c$ beyond their being distinct. We first select $x,y$ to make $x^2+ay^2=m^2.$ If $a=1$ we may use any pythagorean triple, say $(x,y)=(3,4)$ and $m=5.$ Otherwise put $x=a-1,\ y=2,\ m=a+1$ since $(a-1)^2+4a=(a+1)^2.$ [Note: since later we divide by $m^2$ we need $m \neq 0$ so if by chance $a=-1$ we may use $x=5,\ y=4,\ m=3.$] Replacing the first factor of $(1)$ by $m^2$ and multiply through, we have
$$(mu)^2+b(mv)^2=p^2+cq^2.$$
In this, put $q=mv,\ p=ms$ (where $s$ is a "new" variable to be determined) and divide by $m^2$ to obtain
$$u^2+bv^2=s^2+cv^2.\tag{2}$$
At this point the argument branches on whether $|c-b|$ is a square. First assume it is not. Then there are the two rearrangements
$$u^2-(c-b)v^2=s^2,\\ s^2-(b-c)v^2=u^2.\tag{3}$$
Depending on the sign of $c-b$ the left side of one of these is that of a Pell equation $m^2-Dn^2=1$ [with $D>0$ and $D$ not a square], and we may take our right side $s^2$ or $u^2$ to be $1$ to make one of these exactly a Pell equation (with the $1$ on the right). Since the Pell equation has solutions we have a solution to $(1).$
The remaining case, wherein $|c-b|$ is a square, say $k^2,$ is simple; take a 3-4-5 pythagorean triple and multiply it through by $k^2$ depending on which one of $(3)$ is chosen.
Note that, as pointed out by @individ, one really does not need Pell to solve equation $(2).$
A: Never mind: I found a counterexample…
$$(x,a,y,u,b,v,p,c,q)=(1471528, 9777203, 9668027, 5703851, 3364185, 2582134, 16, 819951918602068063799187294877418918951, 5)$$
A: It is necessary to say a few words about the formula about which I have spoken. For the equation: $$(x^2+ay^2)(u^2+bv^2)=p^2+cq^2$$
Need to write this simple formula.
$$x=r^2-as^2$$
$$y=2rs$$
$$u=(k^2-bt^2)(n^2+cj^2)$$
$$v=2kt(n^2+cj^2)$$
$$p=(r^2+as^2)(k^2+bt^2)(n^2-cj^2)$$
$$q=(r^2+as^2)(k^2+bt^2)2nj$$
I think that this formula gives all solutions. Mutually simple solution obtained after reduction to common divisor.
For example there was a similar situation with the equation: $X^2+Y^2=Z^{n}$
It is enough to write the formula generates an endless series of decisions in all degrees.  For this we use the Pythagorean triple.   And the number of their sets.   
$$a^2+b^2=c^2$$   $$a=2ps$$   $$b=p^2-s^2$$   $$c=p^2+s^2$$    
$p,s$ - what some integers.   Then the solution can be written.
$$X=2psc^{n-1}$$   $$Y=(p^2-s^2)c^{n-1}$$   $$Z=c^2$$    
And mutually simple solutions can get if you cut down on common divisor. Although there will be not one simple solution.
