Find XYZ

 Posts: 5
 Joined: Sun Apr 12, 2015 10:17 pm
Find XYZ
$x^2+y^2+z^2=xyz$ , find $(x,y,z)$, where $x,y,z$ are positive odd integers.

 Posts: 5
 Joined: Sun Apr 12, 2015 10:17 pm
Re: Find XYZ
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.
6x'1,6y'1,6z'1
6x'1,6y'1,6z'1

 Posts: 25
 Joined: Sat Feb 07, 2015 5:40 pm
Re: Find XYZ
bhaiya what should i do to solve this problem?
from which sector of math it is?
from which sector of math it is?

 Posts: 5
 Joined: Sun Apr 12, 2015 10:17 pm
Re: Find XYZ
It is from number theory. You should learn modular arithmetic to solve this . Parity will also help.

 Posts: 25
 Joined: Sat Feb 07, 2015 5:40 pm
Re: Find XYZ
here $x = 3, y = 3, z = 3$
$x = y = z = 3$
$3^2 + 3^2 + 3^2 = 27$
$3 ^ 3 = 27$
$x = y = z = 3$
$3^2 + 3^2 + 3^2 = 27$
$3 ^ 3 = 27$

 Posts: 5
 Joined: Sun Apr 12, 2015 10:17 pm
Re: Find XYZ
I wanted the proof that no other solution exists. Otherwise, It's trivial to assume the solution .

 Posts: 5
 Joined: Sun Apr 12, 2015 10:17 pm
Re: Find XYZ
I wanted the proof that no other solution exists. Otherwise, It's trivial to assume the solution .
 seemanta001
 Posts: 13
 Joined: Sat Jun 06, 2015 9:31 am
 Location: Chittagong
Re: Find XYZ
Actually,I think there are infinitely many solutions possible for this equation(!)Taosif Ahsan wrote:I wanted the proof that no other solution exists. Otherwise, It's trivial to assume the solution .
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??
"Sometimes it's the very people who no one imagines anything of who do the things no one can imagine"