hello, one solution is $(x,y,z)=(1;1;1)$. Sonnhard.

Dr Sonnhard Graubner wrote: hello, one solution is $(x,y,z)=(1;1;1)$. Sonnhard. It is unique solution. Obviosly $z>0\to x>0\to y>0$. If $z>1\to y>1$, if $z>1,y>1\to x>1 \to$ contradition. By analogy $z<1$ give contradition.

Here, my solution: $x,y,z>0$ (1) first equation give us: $2(x-y)(x^2+1)+(z-1)^2 +2(z-x)=0(*)$ and $x=min(x,y,z)$ or $x=max(x,y,z)$. (2) second equation give us: $2y^2 (y-1)^2 +y(y-1)^2 +3(y-z)(y^2+1)+(x-1)^2 +4(x-y)=0(**)$ and if $x>y>z$ then it's contradiction. Suppose that, $z>y>x$. Then, from third equation, $2z^5+3y^2+3 = 4xz^2+4x$ $ \Rightarrow $ $ 2z^5+3y^2+3<4xz^2+4x$ $ \Rightarrow $ $2z^5+x^2+1<4xz^2 $ $ \Rightarrow $ $z<x$ and contradiction. Hence, $y=min(x,y,z) $ or $y=max(x,y,z).$ (3) third equation give us: $2z^3(z-1)^2 +4z^2(z-1)^2+2z(z-1)^2+4(z-x)(z^2+1)+3(y-1)^2 +6(y-z)=0(***)$ and if $y>z>x$, then it's contradiction. Suppose that, $x>z>y$. Then, from first equation, $2x^3+z^2+1=2y(x^2+1)$ $ \Rightarrow $ $2x^3+y^2+1<2yx^2+2y$ $ \Rightarrow $ $x<y$ and contradiction. Hence, $z=min(x,y,z)$ or $z=max(x,y,z)$. From (1), (2), (3), we have $x=y$ or $x=z$ or $y=z$. Any the case, from (*),(**),(***), we have a $x=y=z=1$ unique solution. The nice solution by mathuz for ,,Czech-Polish-Slovak Match-2008.''

Rust, i don't understand your solution.

mathuz wrote: Here, my solution: $x,y,z>0$ (1) first equation give us: $2(x-y)(x^2+1)+(z-1)^2 +2(z-x)=0(*)$ and $x=min(x,y,z)$ or $x=max(x,y,z)$. (2) second equation give us: $2y^2 (y-1)^2 +y(y-1)^2 +3(y-z)(y^2+1)+(x-1)^2 +4(x-y)=0(**)$ and if $x>y>z$ then it's contradiction. Suppose that, $z>y>x$. Then, from third equation, $2z^5+3y^2+3 = 4xz^2+4x$ $ \Rightarrow $ $ 2z^5+3y^2+3<4xz^2+4x$ $ \Rightarrow $ $2z^5+x^2+1<4xz^2 $ $ \Rightarrow $ $z<x$ and contradiction. Hence, $y=min(x,y,z) $ or $y=max(x,y,z).$ (3) third equation give us: $2z^3(z-1)^2 +4z^2(z-1)^2+2z(z-1)^2+4(z-x)(z^2+1)+3(y-1)^2 +6(y-z)=0(***)$ and if $y>z>x$, then it's contradiction. Suppose that, $x>z>y$. Then, from first equation, $2x^3+z^2+1=2y(x^2+1)$ $ \Rightarrow $ $2x^3+y^2+1<2yx^2+2y$ $ \Rightarrow $ $x<y$ and contradiction. Hence, $z=min(x,y,z)$ or $z=max(x,y,z)$. From (1), (2), (3), we have $x=y$ or $x=z$ or $y=z$. Any the case, from (*),(**),(***), we have a $x=y=z=1$ unique solution. The nice solution by mathuz for ,,Czech-Polish-Slovak Match-2008.'' I couldn't understand this solution. Can you explain it in more detail? I first find solution $(x, y, z)=(1, 1, 1)$ and change equations like your solution. $2(x-y)(x^2+1)+(z-1)^2+2(z-x)=0, 2y^2(y-1)^2+y(y-1)^2+3(y-z)(y^2+1)+2(x-1)^2+4(x-y)=0, 2z^3(z-1)^2+4z^2(z-1)^2+4(z-x)(z^2+1)+2z(z-1)^2+3(y-1)^2+6(y-z)=0$ I find that in $y \leq x \leq z, z \leq y \leq x, x \leq z \leq y$ these case, there has only $(x, y, z)=(1, 1, 1)$ this solutions. However, I can't prove that this equation hasn't got a solution that isn't (1, 1, 1) in $x \leq y \leq z$ , $z \leq x \leq y$ , $y \leq z \leq x$ this cases. I tried to understand your solution, but I failed.