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$x^2+yz=x$

$y^2+zx=y$

$z^2+xy=z$

I could not do anything to find the solutions. Please give some hints.

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How the tag 'inequality' has come. HINT: subtract any two, –  lab bhattacharjee Aug 25 '13 at 14:45
    
sorry for the tag and thanks for the hint. –  mtm Aug 25 '13 at 14:53
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2 Answers 2

up vote 1 down vote accepted

First assume $x=y=z$. That will give you two solutions.

Otherwise, one of the three unknowns differs from both others. If we assume $x\ne y$ and $x\ne z$, use lab bhatteacharjee's hint to obtain two linear equations in $x,y,z$. You will notice that $y=z$ and $x=1$ follows from these and then there is a unique solution of the original equations (plus two others, by symmetry)

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$x^2-y^2+yz-zx=x-y \Leftrightarrow (x-y)(x+y-z+1)=0$

1.)If $x \neq y,y \ne z,z \ne x$,It becomes a liner equation.It's doable.

2.)If $(x-y)(y-z)(x-z)=0$,Assume $x=y$,It becomes $x(x+z-1)=0,z^2+x^2=z$

2.1)$x=0\Rightarrow y_1=0,z_1=1,y_2=0,z_2=0$

2.2)$x\ne 0\Rightarrow z^2+(1-z)^2=z \Rightarrow z_1=1,x_1=y_1=0,z_2=\frac{1}{2}=x_2$

When $(x-y)(y-z)(x-z)=0$,the answer is $(x,y,z)=(\frac{1}{2},\frac{1}{2},\frac{1}{2}),(0,0,1),(0,1,0),(1,0,0),(0,0,0)$

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Where did you drop $(0,0,0)$? –  Hagen von Eitzen Aug 25 '13 at 14:57
    
Sorry.My mistake.I edited it. –  Smy2012 Aug 25 '13 at 15:00
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