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I have been looking for integer solutions of certain Diophantine equations, one of the simplest examples being

$x^2=4+8y^2+13z^2$.

The ideal answer would be a way to parametrize all the integer equations. It is easy to find infinitely many solutions. For instance, by setting $y=z$, we can use the solutions of Pell's equation. Is there a more general technique to get other solutions? Note that the techniques used for the Legendre's equation do not apply since this equation is not homogeneous.

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    $\begingroup$ One can parametrize all solutions. I recommend the first and last paragraph of my response to a similar question: mathoverflow.net/questions/114707/… $\endgroup$ Commented Apr 20, 2015 at 20:59
  • $\begingroup$ In General, the solution of this equation is determined by the solutions of the Pell equations. Approximately in this form. math.stackexchange.com/questions/74931/… $\endgroup$ Commented Apr 21, 2015 at 4:48
  • $\begingroup$ @GH, Thank you for your answer. Its good to know that it can be solved. I went also further considering the case (not important for my purposes) where the coefficients are $+1$ and $-1$, but not as further as you did. However, this case with coefficients $4, 8, 13$ appears to be more delicate. $\endgroup$ Commented Apr 21, 2015 at 7:18
  • $\begingroup$ Dear @GH, I have looked through your answer to mathoverflow.net/questions/114707/. However, this special case is not quite helpful for me! Could you please kindly post an answer describing an effective algorithm of determination of the orbits of the integral automorphism group of a quadratic form in the set of integer solutions in the general case, say on the example of the equation $x^2-8y^2-13z^2=4$? If necessary, I can learn $p$-adic numbers and adeles in order to understand your answer! $\endgroup$ Commented Apr 21, 2015 at 21:16
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    $\begingroup$ Your problem is a special case of Theorem 6.2 in Chapter 13 of Cassels: Rational quadratic forms. This theorem says that there is an algorithm to find the reduced representations (whose number is finite), and every other representation is an image of a reduced one by an automorph of $-x^2+8y^2+13z^2$. So it remains to find the group of automorphs, for which Sections 4,5,7,8 of Chapter 13 in the book might be helpful. $\endgroup$ Commented Apr 21, 2015 at 21:51

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Let's look at the general case.

$$aX^2+bY^2=cZ^2+N$$

When the coefficients of this equation are not represented in the form of squares. Then a funny pattern emerges. If we know some solution to the equation and we know any solution to the Pell equation. Then we can always construct the following solution to this equation.

$$\left\{\!\begin{aligned} & ax^2+bY^2=cz^2+N \\ & cp^2-as^2=1 \end{aligned}\right. $$

Formally, the relationship between the solvability of the Lagrange equation in general and the solvability of the equivalent Pell equation is obtained.

Numbers can have any signs.

$$f=\sqrt{ac(sz+px)^2+by^2-N}$$

$$\left\{\!\begin{aligned} & X=cxp^2+fs+czps \\ & Z=axps+pf+azs^2 \end{aligned}\right. $$

$$\left\{\!\begin{aligned} & X=cxp^2-2czps+axs^2 \\ & Z=czp^2-2axps+azs^2 \end{aligned}\right. $$

Solutions to the general equation.

$$cp^2-as^2=1$$

They are found using solutions.

$$\left\{\!\begin{aligned} & q^2-can^2=1 \\ & ck^2-at^2=1 \end{aligned}\right.$$

$$\left\{\!\begin{aligned} & p=qk+atn \\ & s=qt+cnk \end{aligned}\right.$$

Consider the example mentioned in the question.

$$8X^2+13Y^2=Z^2-4$$

$X=1 ; Y=1 ; Z=5 $

$a=8 ; b=13 ; c=1$

Solutions to the Pell equation $p^2-8s^2=1$

$(p ; s) - (3 ; 1)$

$$\left\{\!\begin{aligned} & P=3p+8s \\ & S=3s+p \end{aligned}\right. $$

We substitute it and get it. The first gives $(x;Y;z) - (1597;1;4517)$, the second $(x;Y;z)-(443;1;1253)$

For any value of $(X,Y)$, an infinite number of solutions are obtained.

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