Looking modulo $4$ gives that $x,y,z$ are even and by substitution $x=2x_1,y=2y_1,z=2z_1$ you get $x_1^2 + y_1^2 + z_1^2 = 4 x_1 y_1 z_1$. But now all three numbers must again be even and so on leading to something like $x_i^2 + y_i^2 + z_i^2 = 2^{i+1} x_i y_i z_i$ with $x=2^ix_i,y=2^iy_i,z=2^iz_i$ . So there is no solution because every solution must infinitly often be divisible by $2$.

Sorry Zetax, but not all of $x,y,z$ is necessarily even we can have $x$ even , $y$ odd and $z$ odd and the sum of their squares is even !!!!!

But then ($x$ even) the RHS and $x^2$ would be divisible by $4$ and by this also $y^2 + z^2$ and this only works when also $y,z$ are even (since for odd $y$ you have $y^2 \equiv 1 \mod 4$)

what about: $x^2+y^2+z^2=3xyz$ and finally: $x^2+y^2+z^2=kxyz$

_tequilla wrote: $x^2+y^2+z^2=kxyz$ Nice ! I can prove for now that there is no solution for $x,y,z = (0,0,0); (0,0,1)$ but I gave up while trying to prove something for $(1,1,1)$ and $(0,1,1)$ because I don't like to consider too many cases and that was ahead of me when I was trying to solve it. So I'll gladly see the solution

The general formula of this problem is Markov's equation

Markov Equation is the one with $k=3$ [look here http://mathworld.wolfram.com/MarkovEquation.html ]. This one with $k$ is great and probably super-hard or am I wrong ?

_tequilla wrote: what about: $x^2+y^2+z^2=3xyz$ You can easy see that $x=1$, $y=1$ and $z=1$ is a solution. Suppose that $(x,y,z)$ is a solution. If you keep $y$ and $z$ fixed, you will get next quadratic equation. \[ x^2+y^2+z^2=3xyz\Rightarrow x^2-3yzx+y^2+z^2=0 \] Beside $x$ there exist another root $x_1$. Notice that \[ x+x_1=\frac{3yz+\sqrt D}{2}+\frac{3yz-\sqrt D}{2}=3yz, \] which implies that $x_1=3yz-x$ is also an integer. This shows that if $(x,y,z)$ is a solution, $(3yz-x,y,z)$ is also a solution. By induction there are infinitie solutions.

As I remember $k$ can be $1,2,3,4,5,6,8,9$ So what is the solution for the other cases? (for $k=4,5,6,8,9$)

Let $x,y,z$ must be all even to satisfy the condition Bcoz if taken even and odd it violates $x=2a_1,y=2k_1,z=2f_1$ .,$a_1,k_1,f_1\in\mathbb{N} a_1,a_2,a_3 \ge 1$ $(a_1)^2+(k_1)^2+(f_1)^2=4a_1k_1f_1$ Further $a_1=2a_2,n=2p_2,p=2c_2, v(a)\ge 1$ $(a_2)^2+(p_2)^2+(c_2)^2=8a_2p_2c_2$ Looking at this we conclude $(x_i)^2+(y_i)^2+(z_i)^2=2^{i+1}x_iy_iz_i$ From there it follows and it is true for four variables $x, y, z, x_n$ $a^4+b^4+c^4+d^4=2abcd$ Same way solutions are taken. $parity$ violates if we take $(o, e, o), (o, o, o), (e, o, e),... $

This is just pell's equation infinite descent!