For students of class 9-10 (age 14-16)
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,
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,$$3y'z'=x'+b$$.
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??
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