Find all real triples $(x,y,z)$ such that: $x^2+1=2y,y^2+1=2z,z^2+1=2x$

Question

Find all real triples $(x,y,z)$ such that: $x^2+1=2y,y^2+1=2z,z^2+1=2x$.

Can it be solved avoiding equations of orders higher than 2?

Answer 1

Adding all the equations, we get -

$$x^2+1+y^2+1+z^2+1=2x+2y+2z$$

$$x^2-2x+1+y^2-2y+1+z^2-2z+1=0$$

$$(x-1)^2+(y-1)^2+(z-1)^2=0$$

$$\implies x=y=z=1$$

Answer 2

HINT: we get $$x,y,z\geq 0$$ eliminating $y,z$ from the given System we get $$- \left( {x}^{6}+2\,{x}^{5}+7\,{x}^{4}+12\,{x}^{3}+31\,{x}^{2}+50\,x+ 89 \right) \left( x-1 \right) ^{2} =0$$ therefore we obtain $$x=y=z=1$$ the plonomial above has only complex solutions