Does the Diophantine equation $(x^2+ay^2)(u^2+bv^2) = p^2+cq^2$ admit a complete solution? In this MSE question/thread, I have been discussing the equation
$$
(x^2+ay^2)(u^2+bv^2) = p^2+cq^2,  \tag{$\star$}
$$
where $x,a,y,u,b,v,p,c,q$ are integers. I posed a conjecture which turned out to be false. Now, as an alternative, I'm looking for a "complete integer solution" (a.k.a. integer parameterization); MO seemed like a better place to look for one, since the issue of the composition of binary quadratic forms is notoriously challenging.
Is a complete solution or parameterization already known?
If not, what's the right way to go about deriving one? I can imagine trying to resolve $x^2+ay^2=mn^2$, where [hopefully] $m$ is squarefree; doing the same for $u^2+bv^2=jk^2$ would then leave $p^2+cq^2=jm(kn)^2 = j'm'(\delta kn)^2$ where $\delta=\gcd(j,m)$, which could be resolved the same way. Is that likely the best approach?
The main application is the solution of higher-degree equations which have been reduced to the form ($\star$).
 A: I don't think it has been explicitly mentioned yet, but presumably you are interested in integer solutions for given values of $a, b, c$.
L E Dickson's "History of the Theory of Numbers" vol 2 [ http://books.google.co.uk/books?id=eNjKEBLt_tQC ] page 433 details necessary and sufficient conditions for a homogenous quaternary quadratic $a x^2 + b y^2 + c z^2 + d t^2 = 0$ to have non-zero integer solutions, for a given set of non-zero integers $a, b, c, d$.
So if you can find numerical integer values of $x, y$ such that, denoting $A := x^2 + a y^2$, $A u^2 + A b v^2 - p^2 - c q^2 = 0$ has no non-zero integer solutions $u, v, p, q$, then that would rule out a general parametrisation, at least for unrestricted values of $a, b, c$. I suspect there are such numeric values of $A, b, c$.
If on the other hand $a, b, c$ are also thrown into the pot, i.e. as variables on the same footing as $x, y$ etc, then it is trivial to find every integer solution by simply taking $a, b, p, q = x A, z B, x z P, x z Q$ respectively to obtain an equation (in general *) bilinear in x, z which can then be expressed in the form $(x + F)(z + G) = H$, in which $F, G, H$ do not depend on $x, z$, and then one can simply take $x + F, z + G = H / t, t$ respectively for some parameter $t$ and clear denominators.
(*) In the special case when $P^2 + c Q^2$ = 1 we have an equation linear in $x, z$. But in that case we can take $x, z = y^2 X, t^2 Z$ respectively to obtain $(A + X)(B + Z) = X Z$, and solve this as above for $A, B$ to obtain $A + X, B + Z = X/t, Z t$ respectively.
A: I think that this method of calculation it is necessary to separately draw.
As I have repeatedly said formula in General looks pretty bulky.  And still remain questions about the completeness of the solution. So I decided that solutions should be found a little differently.
In Diofantos equation:
$$(x^2+ay^2)(u^2+bv^2)=p^2+cq^2$$
Put some numbers:    $t,y$ 
Decompose to factor the following expression:  $ct^2-ay^2=AB$
Then we can define the following numbers:
$$s=\frac{A-B}{2}+t$$
$$x=\frac{A+B}{2}$$
Next, you can specify the desired number: $v$
Subject to the following expression for the multiplier: $cs^2-bv^2=FJ$
This will allow us to unambiguously identify numbers:
$$k=\frac{J-F}{2}+s$$
$$u=\frac{F+J}{2}$$
And for the full solution will be found by the formula two other numbers.
$$p=ks+(c+1)ts-tk-s^2$$
$$q=s^2-tk$$
A: As posted in my comment above, the case $a=b=c=1$ is relatively trivial to solve, using existing (nearly "classical") solutions to the 2.2.4 Diophantine sums-of-squares equation $$X_1^2 + X_2^2 = Y_1^2+Y_2^2+Y_3^2+Y_4^2.$$
Since nobody else was able or willing to develop a complete solution for the more general equation in the title, I am working on it myself. The complete solution will be complicated — as fully expected/anticipated — but is within reach. I will post it back here when I've fully verified my proof/algorithm.
A: For the equation:
$$(x^2+ay^2)(u^2+bv^2)=r^2+cq^2$$
This solution will suit you?
$$x=cp^2+k^2-at^2$$
$$y=2kt$$
$$u=cp^2+k^2-bs^2$$
$$v=2ks$$
$$r=(bs^2+k^2-cp^2)(at^2+k^2-cp^2)-4ck^2p^2$$
$$q=2kp(at^2+bs^2+2k^2-2cp^2)$$
$t,p,k,s$ - integers of any sign.
A: Still not clear what you want.  If you need a method, the method of finding solutions I have already in a previous post wrote.  If regarding the parameterization.
Then for the equation:
$$(x^2+ay^2)(u^2+bv^2)=z^2+cr^2$$
You can  write many upgrades parameterization similar to the previous one.
$$x=cq^2-an^2+p^2$$
$$y=2pn$$
$$u=ck^2-bs^2+t^2$$
$$v=2ts$$
$$z=(p^2+an^2-cq^2)(t^2+bs^2-ck^2)-4cpqtk$$
$$r=2tk(p^2+an^2-cq^2)+2pq(t^2+bs^2-ck^2)$$
$q,n,p,k,s,t$ - integers asked us.
If this formula is not satisfied, then you can write in another form. But the point remains the same.  Decisions revolve around the same equation.
If such a decision is not satisfied - tell.  
A: Once these equations are referred to - it is necessary to write and solutions.
For the equation:
$$X_1^2+X_2^2=Y_1^2+Y_2^2+Y_3^2$$
Solutions have the form:
$$X_1=t^2+2(p+s-k)t+2k^2+2p^2+4ps-4pk-2sk$$
$$X_2=t^2+2(p+s-k)t+2k^2+2s^2+4ps-2pk-4sk$$
$$Y_1=t^2+2(p+s-k)t+2k^2+2ps-2pk-2sk$$
$$Y_2=t^2+2(p+s-k)t+2ps$$
$$Y_3=2(p+s-k)(t+p+s-k)$$
For the equation:
$$X_1^2+X_2^2=Y_1^2+Y_2^2+Y_3^2+Y_4^2$$
Solutions have the form:
$$X_1=t^2+2(p+s-y)t+k^2+2y^2+2p^2-4yp-2ys+4ps$$
$$X_2=t^2+2(p+s-y)t+k^2+2y^2+2s^2-2yp-4ys+4ps$$
$$Y_1=t^2+2(p+s-y)t+k^2+2y^2-2yp-2ys+2ps$$
$$Y_2=2(p+s-y)k$$
$$Y_3=2(p+s-y)(t+p+s-y)$$
$$Y_4=t^2+2(p+s-y)t+k^2+2ps$$
$k,y,t,p,s$ - integers asked us.
