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A friend asked me the following question several days ago, and we still do not have a solution.

Prove that the system of equations below has only the solution $(x, y, z)=(1, 1, 1)$. $$ \begin{cases} x+y^2+z^3=3\\ y+z^2+x^3=3\\ z+x^2+y^3=3 \end{cases} $$

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    $\begingroup$ What rihkddd mentioned. I tried to manipulate it into an equation of the form $(y−x)^2A+(z−y)^2B+(x−z)^2C=0$ where $A,B,C$ are perfect squares, to show symmetry. I got $(y−x)A+(z−y)B+(x−z)C=0$, but that is not exactly what I am looking for. $\endgroup$
    – Emre
    Feb 18, 2014 at 19:15

2 Answers 2

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The question should be wrong since at least it can be checked by Wolfram Alpha that it has more than one group of solution (in fact it has $27$ groups of solution).

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  • $\begingroup$ I forgot to mention that we are working with real solutions. $\endgroup$
    – Emre
    Feb 23, 2014 at 20:32
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maybe you could find that x,y,z are symmetyical ,which means x,y,z have no difference with each other ,so x=y=z.you can solve this original equation by transform to x+x^2+x^3=3,so x =1,the so the only solution is (x,y,z)=(1,1,1). I'm not native English spaker ,sorry for my bad English,hope you can understand what I mean.

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  • $\begingroup$ Yes, under the hypothesis that $x=y=z$ you can easily show that the only solution is $x=y=z=1$. However, the true question is whether there're other, non-symmetrical solutions. $\endgroup$
    – TZakrevskiy
    Feb 18, 2014 at 15:00

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