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$$x^2 + y^2 + z^2 = 3xyz$$

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.

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Your suggestions do not solve the equations. If $x=0$ then necessarily $y=z=0$. – Hagen von Eitzen Dec 9 '12 at 12:42
haha I only realised my mistake now, i was solving a different equation, apologies. so would the only solutions for this be x = y = z = 0 and 1 – fosho Dec 9 '12 at 12:46
There are infinitely many solutions which you can generated by a procedure. see also – user51427 Dec 9 '12 at 20:00
This question may be helpful:… – Adam Bailey Dec 10 '12 at 14:41
up vote 0 down vote accepted

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. $(5, 29, 433)$.

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does ordered triplet mean x≤y≤z – fosho Dec 9 '12 at 12:50
Hm, if I wanted to identify $(1,2,5)$ with $(5,1,2)$, I'd say I count unordered triples and if I count them as different solutions I'd say I count ordered triples. Of course the former implicitly allows me to impose an order to my liking on the triple, so that may be confusing. – Hagen von Eitzen Dec 9 '12 at 12:54

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