SAUDI ARABIAN MATHEMATICAL COMPETITIONS

From the given conditions, we have $$xz=2z-y,xy=2x-z,yz=2y-x.$$ Taking the sum of these equations, side by side, we have $$xy+yz+zx=x+y+z=T.$$ From $xz=2z-y$, we also can get $2z-x-y=xz-x\to 3z-T=x(z-1)$. Make the similar equations and multiply them, we get $$xyz(x-1)(y-1)(z-1)=(3x-T)(3y-T)(3z-T).$$ Note that $(x-1)(y-1)(z-1)=xyz-(xy+yz+zx)+(x+y+z)-1=xyz-1$ and $$\begin{array}{rl}(3x-T)(3y-T)(3z-T)& =27xyz-9T(xy+yz+zx)+T^2(3x+3y+3z)-T^3\\ & =27xyz-9T^2+2T^3\end{array}$$ Thus $$\begin{array}{c}xyz(xyz-1)=27xyz-9T^2+2T^3\text{ or }(xyz{)}^2-28xyz=2T^3-9T^2.\end{array}\text{\hspace{1em}(1)}$$ By the similar transformation, $xz=2z+y\to xz+2x=2z+2x-y\to x(z+2)=2T-3y$. Make the similar equations and multiply them, side by side, we get $$(xyz{)}^2+6(xyz)T+35xyz=18T^2-4T^3.$$ On the other hand, $xz=2z-y\to xyz=2yz-y^2$. Make the similar equations and multiply them, side by side, we get $3xyz=2(xy+yz+zx)-(x^2+y^2+z^2)=4(xy+yz+zx)-(x+y+z{)}^2=4T-T^2$. So we have $xyz=\frac{4T-T^2}{3}$, then by substituting to equation (1), we get $$\begin{array}{c}{\left(\frac{4T-T^2}{3}\right)}^2-28\left(\frac{4T-T^2}{3}\right)=2T^3-9T^2\end{array}\text{\hspace{1em}(2)}$$ Solving this equation, we have $T\in \{0,3,7,16\}$.

1. If $T=0$, then $xyz=0$, a contradiction. 2. If $T=16$, then $xyz=-64$ and this pair $(T,xyz)=(16,-64)$ does not satisfy equation (2). 3. If $T=3$, then $xyz=1$ and by Vieta's theorem, $x,y,z$ are real roots of $t^3-3t^2+3t-1=0$, which implies that $x=y=z=1$. 4. If $T=7$, we get $xyz=-7$ and $x,y,z$ are roots of $t^3-7t^2+7t+7=0$, this also have three distinct real roots.

Hence, all possible values of $T$ are $3$ or $7$.