Integer solutions of the equation: $x^2+y^2+z^2=kxyz$

Question

Given the equation: $$x^2+y^2+z^2=kxyz$$ with: $(k,x,y,z)\in\mathbb{N}$, the only solution for $k=2$ is: $x=0,y=0,z=0$. For what values of $k$ the equations has solutions in which $x,y,z$ are different from zero? Thanks.

Answer 1

For $k=3$ this is the Markov Equation. It has the solution $(1,1,1)$ and from this solution you can build any other by the identification $$(x_0,y_0,z_0) \to (x_0, y_0, 3x_0y_0 - z_0)$$ and by noticing you can permute $x,y,z$.

$$(1,1,1) \to (1,1,2)$$

$$(1,2,1) \to (1,2,5)$$

$$(1,5,2) \to (1,5,13)$$

$$(2,5,1) \to (2,5,29)$$ etc.

Answer 2

You must write the formula, without understanding all the same will not happen!

To show an example. If you will ask is for example $z=1$ You can write infinitely many solutions.

Even such: $$(1,89,233)$$ $$(1,1597,4181)$$ $$(1,28657,75025)$$ Need to write a formula in General.

For the equation: $$x^2+y^2+z^2=qxyz$$

The solution, you can write. $t$ - what some integer.

Then using the solutions of the equation Pell. $p^2-(t^2+4)s^2=\pm1$

The solutions have the form:

$$z=\frac{t^2+2}{q}$$

$$x=\frac{(t^2+2)(p^2-2tps+(t^2+4)s^2)}{qt}$$

$$y=\frac{(t^2+2)(p^2+2tps+(t^2+4)s^2)}{qt}$$

$q,t$ - it is necessary to choose such that decisions were intact.