<s>[INST] If $x,y,z \in \mathbb{R}$ are such that $x^2+x=y^2$, $y^2+y=z^2$ and $z^2+z=x^2$, prove that $(x-y)(y-z)(z-x)=1.$ If $x,y,z$ are real numbers that not equal to 0 such that $x^2+x=y^2$, $y^2+y=z^2$ and $z^2+z=x^2$ prove that $(x-y)(y-z)(z-x)=1$.
I solve this problem by adding each two equations, but I'm interested can we solve this problem by subtracting equations? By subtracting we get $(x-y)(x+y+1)=(y-z)(y+z)$.
 [/INST] Adding the three equations we get
$$x+y+z=0.$$
So then, if we rewrite the equations as $y^2-x^2=x$, $z^2-y^2=y$, etc., we find $(y+x)(y-x)=x$, $(z+y)(z-y)=y$, $(x+z)(x-z)=z$, which in view of the above relation we can rewrite as $(x-y)z=x$, $(y-z)x=y$, $(z-x)y=z$. Multiplying these three equations, and using that $xyz \neq 0$, we get the desired conclusion.
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