Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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.

share|improve this question
    
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
1  
There are infinitely many solutions which you can generated by a procedure. mathworld.wolfram.com/MarkovEquation.html see also mathworld.wolfram.com/MarkovNumber.html –  user51427 Dec 9 '12 at 20:00
    
This question may be helpful: math.stackexchange.com/questions/94394/… –  Adam Bailey Dec 10 '12 at 14:41
add comment

1 Answer

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. $(5, 29, 433)$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.