Solving quadratic Diophantine $a x^2 + b y^2 = c z^2$ given $a,b,c$

Question

I do not have much background in Diophantine equations, but I'm curious about how to find families of solutions to them, particular to equations that are quadratic in the variables. I'm trying to understand whether there are limitations to a procedure I've seen in a few locations on MSE, in that it might not provide a complete set of solutions.

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It seems like there is a nice procedure for constructing solutions to equations of the type $$a x^2 + b y^2 = c z^2$$ when a solution exists. The recipe appears to be to consider rewrite the equation in terms of rationals $r,s$ so that $$a r^2 + b s^2 = c.$$ Given a solution to the original equation, we get a solution $(r_0, s_0)$ to the second equation. Then one solves the pair of equations $$a r^2 + b s^2 = c$$ $$s = s_0 + m(r-r_0)$$ in terms of the rational number $m$, giving $r$ and $s$ as a rational function of $m$.

For example, up to a change in notation, Will Jagy answered the question of finding solutions to $x^2+y^2 = 5 z^2$ by noting that $x=1,y=2$ is a solution. Following the steps above, he found $$r = \frac{2 m^2 - 2 m - 2}{m^2 + 1}$$ $$s = \frac{- m^2 - 4 m + 1}{m^2 + 1}.$$ Writing $m=\frac{p}{q}$, this gives $$r = \frac{2 p^2 - 2 pq - 2q^2}{p^2 + q^2}$$ $$s = \frac{- p^2 - 4 pq + q^2}{p^2 + q^2}$$ from which one can deduce that the set $$(x,y,z) = (2 p^2 - 2 pq - 2q^2, - p^2 - 4 pq + q^2, p^2 + q^2)$$ solves the original equation of $x^2 + y^2 = 5 z^2$ for every $p,q \in \mathbb{Z}$.

However, I'm having a hard time convincing myself that these are indeed all the solutions.

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My question in particular is on $x^2 + 68 y^2 = z^2$. A little playing around gets the solutions $(x,y,z) = (1,4,33)$.

When I follow the procedure, I get

$$r = \frac{68 m^2-544 m-1}{33(1+68 m^2)}$$ $$s = -\frac{2 \left(136 m^2+m-2\right)}{33 \left(68 m^2+1\right)}$$ which to me signals that $$(x,y,z) = (-68 p^2 + 544 p q + q^2,2 (136 p^2 + pq - 2 q^2), 33 (68 p^2 + q^2))$$ solve the equation $x^2 + 68 y^2 = z^2$.

However, I again fear that this is not giving all solutions to the Diophantine equation. For example, I fear it's missing $(x,y,z) = (8,3,26)$.

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Can I use the technique above or a modification of it to find the complete set of solutions to a given Diophantine equation $a x^2 + b y^2 = c z^2$ (specifically, one that has solutions and for which I know a particular solution)? For example, how should I modify or supplement the solution above for $x^2 + 68 y^2 = z^2$ to get a complete set of solutions?

Answer 1

See how well these three recipes do. If i have time I write a program that finds all solutions by brute force, call those "raw," then compare with the recipes. Easy to miss comparisons by hand. Since you have $x^2 + 68 y^2 = z^2$ we may freely use absolute value signs.

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

A:

$$ x = |4u^2 - 17 v^2| \; , \; \; y = |2uv| \; , \; \; z = 4 u^2 + 17 v^2 $$

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

B:

$$ x = |4u^2 +4uv - 16 v^2| \; , \; \; y = |2uv + v^2| \; , \; \; z = 4 u^2 +4uv + 18 v^2 $$

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

C:

$$ x = |u^2 - 68 v^2| \; , \; \; y = |2uv | \; , \; \; z = u^2 + 68 v^2 $$

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

Comparison program:

   1       0       1     C         1       0
   1       0       1       RAW   
   1       4      33     A         2       1
   1       4      33     A         2      -1
   1       4      33       RAW   
   8       3      26     B         1       1
   8       3      26     B         2      -1
   8       3      26       RAW   
   8       5      42     B         2       1
   8       5      42     B         3      -1
   8       5      42       RAW   
  13      18     149     C         9       1
  13      18     149     C         9      -1
  13      18     149       RAW   
  13       2      21     A         1       1
  13       2      21     A         1      -1
  13       2      21       RAW   
  16     105     866     B        13      -5
  16     105     866     B         8       5
  16     105     866       RAW   
  16       1      18     B         0       1
  16       1      18     B         0      -1
  16       1      18     B         1      -1
  16       1      18       RAW   
  16      39     322     B         5       3
  16      39     322     B         8      -3
  16      39     322       RAW   
  19      14     117     C         7       1
  19      14     117     C         7      -1
  19      14     117       RAW   
  19       6      53     A         3       1
  19       6      53     A         3      -1
  19       6      53       RAW   

This is a page from Mordell, Diophantine Equations.