$$x^2 + y^2 + z^2 = 3xyz$$
How many ordered triples $(x,y,z)$ are there that satisfy the above equation.
are the only solutions $x=y=z=0$ and $1$?
Are there non trivial solutions?
I saw this problem in a friends textbook but cannot remember the name of it therefore cannot cite the the exact source.
There are $41$ solutions (this refers to the original problem statement that included the condition $-10<x,y,z<10$). If we additionally require $x\le y\le z$, there are only the following $10$: $$[-5, -2, 1]\\ [-5, -1, 2]\\ [-2, -1, 1]\\ [-2, -1, 5]\\ [-1, -1, 1]\\ [-1, -1, 2]\\ [0, 0, 0]\\ [1, 1, 1]\\ [1, 1, 2]\\ [1, 2, 5]$$ Without the restriction $|x|,|y|,|z|<10$, a whole bunch of additional solutions comes up, e.g. $(5, 29, 433)$.