BdMO Online Forum

$x^2+y^2+z^2=xyz$ , find $(x,y,z)$, where $x,y,z$ are positive odd integers.

x = 2, y = 3, z = 4

bhai give the solution

x'^2+y'^2+z'^2=3x'y'z' , if you can solve this ,then we are done.Here is some informations that might help.
6|x'-1,6|y'-1,6|z'-1

bhaiya what should i do to solve this problem?
from which sector of math it is?

It is from number theory. You should learn modular arithmetic to solve this . Parity will also help.

here $x = 3, y = 3, z = 3$
$x = y = z = 3$
$3^2 + 3^2 + 3^2 = 27$
$3 ^ 3 = 27$

I wanted the proof that no other solution exists. Otherwise, It's trivial to assume the solution .

I wanted the proof that no other solution exists. Otherwise, It's trivial to assume the solution .

Taosif Ahsan wrote:I wanted the proof that no other solution exists. Otherwise, It's trivial to assume the solution .

Actually,I think there are infinitely many solutions possible for this equation(!)
Clearly, $$x^2+y^2+z^2=xyz$$ modulo $3$ shows that they are all multiples of $3$.Now, we can write $$x=3x'$$,$$y=3y'$$ and $$z=3z'$$.
This leads us to $$x'^2+y'^2+z'^2=3x'y'z'$$, which is "Markov's Equation".Clearly, here $$(x,y,z)=(1,1,1)$$ is a solution.
Now in this equation,we assume $x$ is the root.We can rewrite it as,
$$a^2+(3y'z')a+(y'+z')=0$$ (i)
then by using the quadratic equation formula,we can assume b is the other root.Now we can observe that,
$$(a - x')(a - b)=a^2-(x'+b)a+x'b$$ (ii)
Comparing (i) and (ii) we get,$$3y'z'=x'+b$$.
Thus,$$b=3y'z'-x'$$.
Interchanging $x$,$y$ and $z$, we can write them as $(3y'z'-x',y',z')$ and $(x',3y'z'-x,z')$ and $(x',y',3y'z'-x')$.
Now from the first solution $(1,1,1)$, we can generate $(2,1,1)$,$(2,1,5)$,$(2,29,5)$ and infintely many solutions(!)
Thus, $$x^2+y^2+z^2=xyz$$ will have solutions simply 3 times a solution of $$x'^2+y'^2+z'^2=3x'y'z'$$.Which are $(3,3,3)$,$(6,3,3)$,$(6,3,15)$,$(6,58,15)$ and infinitely many more.
But I can't really prove that Markov triples are infinitive.Can anybody help me with that??