parmenides51 wrote: Determine all triples (x, y, z) of non-positive real numbers that satisfy the following system of equations $\begin{cases} x^2 - y = (z - 1)^2\\ y^2 - z = (x - 1)^2 \\ z^2 - x = (y -1)^2 \end{cases}$. Comment. If $x,y$, and $z$ are non-positive real numbers, then $x+y+z\le0$. But the original system of equations yields $$x^2 - y+ y^2 - z + z^2 - x = (z - 1)^2+(x - 1)^2 + (y -1)^2,$$leading to $x+y+z=3>0$, a contradiction.

However, if we drop all domain restrictions and assume instead solving in complex numbers, then we find that it has four solutions and all of them happen to be real. My analysis furthermore shows that in the non-trivial solutions the variables should be expressible as $\sum_{k=1}^na_k\cos q_k\pi$ where $a_k,q_k\in\mathbb{Q}$.

Let $x=a+1,\ y=b+1,\ z=c+1.$ Then the system is equivalent to $a^2+2a-c^2=b\ (1), \ \ b^2+2b-a^2=c\ (2),\ \ c^2+2c-b^2=a.\ $ Adding we get $a+b+c=0,$ and therefore $c=-(a+b). \ (3)$ Substituting $(3)$ into $(1)$ we get $a=\frac{b(b+1)}{2(1-b)}\ (4),$ (where clearly $b\ne 1$). Now substituting $(3)$ and $(4)$ into $(2)$ we get $b(3b^3-21b+14)=0$. The first solution is $b=0,$ which leads to $a=b=c=0,$ or $x=y=z=1.$ The solutions of the cubic are given by $b=2\sqrt{\frac{7}{3}}\cos\left[\frac{1}{3}\arccos\left(-\sqrt{\frac{3}{7}}\right)-\frac{2\pi k}{3}\right],$ where $k=0,1,2,$ and they are all real solutions. In approximate form, the solutions are $(a,b,c)=(-2.93122,2.21124,0.71998)$ and its cyclic permutations.

parmenides51 wrote: Determine all triples (x, y, z) of non-positive real numbers that satisfy the following system of equations $\begin{cases} x^2 - y = (z - 1)^2\\ y^2 - z = (x - 1)^2 \\ z^2 - x = (y -1)^2 \end{cases}$. the official english translation of the solutions gives $x=y=z=1$ (which is against non-positive) I look in the original Dutch and it was written not-negative instead of not-positive I shall correct it

parmenides51 wrote: Determine all triples (x, y, z) of non-negative real numbers that satisfy the following system of equations $\begin{cases} x^2 - y = (z - 1)^2\\ y^2 - z = (x - 1)^2 \\ z^2 - x = (y -1)^2 \end{cases}$. In this case the solution is straightforward. If $x\in [0,1),$ then from the first expression we get $y\in [0,1)$ and from the second expression we get $z\in [0,1).$ However, this contradicts the fact that adding the three expressions we get $x+y+z=1.$ Similarly, if $x\ge 1,$ the third expression gives $z\ge 1,$ and then the second expression gives $y\ge 1;$ now, because $x+y+z=3,$ the equalities work only if $x=y=z=1.$

And here are the four solution tuples $(x,y,z)$ (the non-negative constraint dropped). $(x,y,z)\in\{(1,1,1)$, $(a,b,c)$, $(b,c,a)$, $(c,a,b)\}$ $a=1+2\cos\frac{2\pi}{63}-2\cos\frac{10\pi}{63}+\frac{2}{3}\cos\frac{2\pi}{9}+2\cos\frac{16\pi}{63}-2\cos\frac{46\pi}{63}+\frac{4}{3}\cos\frac{8\pi}{9}$ $b=1+2\cos\frac{40\pi}{63}-2\cos\frac{52\pi}{63}+\frac{2}{3}\cos\frac{4\pi}{9}+2\cos\frac{58\pi}{63}-2\cos\frac{38\pi}{63}+\frac{4}{3}\cos\frac{2\pi}{9}$ $c=1+2\cos\frac{44\pi}{63}-2\cos\frac{32\pi}{63}+\frac{2}{3}\cos\frac{8\pi}{9}+2\cos\frac{26\pi}{63}-2\cos\frac{4\pi}{63}+\frac{4}{3}\cos\frac{4\pi}{9}$