<s>[INST] Find all real triples $(x,y,z)$ such that: $x^2+1=2y,y^2+1=2z,z^2+1=2x$ 
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?
 [/INST] Adding all the equations, we get -
$$x^2+1+y^2+1+z^2+1=2x+2y+2z$$
$$x^2-2x+1+y^2-2y+1+z^2-2z+1=0$$
$$(x-1)^2+(y-1)^2+(z-1)^2=0$$
$$\implies x=y=z=1$$
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