any ideas?

The system is symmetric so let $x \geq y \geq z$. We have $x^2=yz+y-z+1...(1)$ $y^2=xz+x-z+1...(2)$ $z^2=xy+x-y+1...(3)$ \[(2)-(3)=y^2-z^2=x(z-y)+(y-z)\]\[(y-z)(y+z)=(y-z)(1-x)\]$y=z$ or $x+y+z=1$. $i)y=z$ $x^2=y^2+1$ and $y^2=xy+x-y+1$. $\implies x^2-y^2=(y^2+1)-(xy+x-y+1)=y(y-x)+(y-x)$ $\implies (x-y)(x+y)=(y+1)(y-x)$ If $x=y=z$ then we get $0=1$ from $(1)$ so we can divide both side by $(x-y)$. $\implies x+y=-y-1$ $\implies \boxed{x=-2y-1}$ We have $x^2=y^2+1$ so \[(2y+1)^2=y^2+1\]\[3y^2+4y=0\]\[y(3y+4)=0\]$y=z=0,x=1$ is not a solution so $y=z=-\frac{4}{3},x=\frac{5}{3}$ is the uniqe solution for this case. $ii)x+y+z=1$ From $(1)$ we get \[x^2=y(1-x-y)+y+(x+y)=-xy-y^2+3y+x\]$\implies x^2+y^2=-xy+3y+x$ From $(2)$ we get \[y^2=x(1-x-y)+x+(x+y)=-xy-x^2+3x+y\]$\implies x^2+y^2=-xy+3x+y$ $\implies -xy+3y+x=x^2+y^2=-xy+3x+y$ $\implies x=y$ From $(3)$ we get \[(1-2x)^2=x^2+1\]\[4x^2-4x+1=x^2+1\]\[3x^2=4x\]If $x=0$ then $z=1$ but we have $x \geq z$ so $x=y=\frac{4}{3},z=-\frac{5}{3}$ Answer:$(x,y,z)=(\frac{5}{3},-\frac{4}{3},-\frac{4}{3}),(\frac{4}{3},\frac{4}{3},-\frac{5}{3})$ and their permutations.

bin_sherlo wrote: The system is symmetric so let $x \geq y \geq z$. We have $x^2=yz+y-z+1...(1)$ $y^2=xz+x-z+1...(2)$ $z^2=xy+x-y+1...(3)$ \[(2)-(3)=y^2-z^2=x(z-y)+(y-z)\]\[(y-z)(y+z)=(y-z)(1-x)\]$y=z$ or $x+y+z=1$. $i)y=z$ $x^2=y^2+1$ and $y^2=xy+x-y+1$. $\implies x^2-y^2=(y^2+1)-(xy+x-y+1)=y(y-x)+(y-x)$ $\implies (x-y)(x+y)=(y+1)(y-x)$ If $x=y=z$ then we get $0=1$ from $(1)$ so we can divide both side by $(x-y)$. $\implies x+y=-y-1$ $\implies \boxed{x=-2y-1}$ We have $x^2=y^2+1$ so \[(2y+1)^2=y^2+1\]\[3y^2+4y=0\]\[y(3y+4)=0\]$y=z=0,x=1$ and $y=z=-\frac{4}{3},x=\frac{5}{3}$ are solutions for this case. $ii)x+y+z=1$ From $(1)$ we get \[x^2=y(1-x-y)+y+(x+y)=-xy-y^2+3y+x\]$\implies x^2+y^2=-xy+3y+x$ From $(2)$ we get \[y^2=x(1-x-y)+x+(x+y)=-xy-x^2+3x+y\]$\implies x^2+y^2=-xy+3x+y$ $\implies -xy+3y+x=x^2+y^2=-xy+3x+y$ $\implies x=y$ From $(3)$ we get \[(1-2x)^2=x^2+1\]\[4x^2-4x+1=x^2+1\]\[3x^2=4x\]If $x=0$ then $z=1$ but we have $x \geq z$ so $x=y=\frac{4}{3},z=-\frac{5}{3}$ Answer:$(x,y,z)=(\frac{5}{3},-\frac{4}{3},-\frac{4}{3}),(1,0,0),(\frac{4}{3},\frac{4}{3},-\frac{5}{3})$ and their permutations. I think you have a mistake (1,0,0) isn't a solution