Solutions to $ax^2 + by^2 = cz^2$ The integer solutions to the equation $x^2 + y^2 = z^2$ are very well studied. I'm wondering if there's any literature about the integer solutions to the equation $ax^2 + by^2 = cz^2$ where a,b,c are non-zero integers. Here are some of my questions:


Given a,b,c, is there an explicit way of writing all the solutions? 
If not, could you generate all the solutions given one of the solutions?
If this is the case, are there any algorithms that are designed to look for one of these solutions?
Are there any additional conditions that these solutions must satisfy? 
What happens when you're only given a and b?
Proof of Legendre&#39;s theorem on the ternary quadratic form 

I like the discussion of Legendre's theorem in Cassels, Rational Quadratic Forms, Theorem 4.1, pages 80-82, as he is careful about pointing out that certain products must be quadratic residues for each prime dividing each entry $(a,b,c).$

Next, if $a,b,c$ are positive, and there is at least one nontrivial solution $(x_0,y_0,z_0)$ then we may find all rational points on the ellipse $a \alpha^2 + b \beta^2 = c$ by drawing all lines of rational slope through $(x_0/z_0, y_0/z_0).$ For each such line, there is a second rational point on the ellipse, with the slope $m$ as a parameter. Multiplying back through by the denominators gives a new integral solution to $a x^2 + b y^2 = c z^2.$ It is exactly this method that gives primitive Pythagorean triples $(2rs, s^2 - r^2, s^2 + r^2).$   I've done that a few times, i will try to find some of  my answers on this site.

Found one anyway, Generating Pythagorean triples for $a^2+b^2=5c^2$? 

Found another: note that more than one (two, here) recipe may be required to get all solutions. Pythagorean triples with additional parameters

Oh, for this few variables it is good enough to take lines with rational slope, but in general, such as $a x^2 + b y^2 + c z^2 = d w^2,$ it is better to find some rational point on the ellipsoid $(x_0/w_0, y_0/w_0, z_0/w_0)$ and then use a parameter $t$ and integers, call them $p,q,r,$ and find the other ellipsoid point 
$(x_0/w_0 + pt, y_0/w_0 + qt, z_0/w_0 +rt), $ where we can then eliminate $t.$
Above mentioned equation is shown below:

$ax^2 + by^2 = cz^2$

The above equation has parametric solution for $(a,b,c)=(35,10,3)$

$x=5k^2-20k-4$

$y=10k^2+20k-32$

$z=25k^2-20k+60$

For $k=1$ we get:

$35(19)^2+10(2)^2=3(65)^2$