Just a quick question. I did the question as well but I started with the $ \text{Rational Roots Theorem}$ and got $ x=-1$ as a root. Etc... I am just curious how you got the $ x=y=z= \pm \sqrt{ \frac{3}{2}}+1$ from the polynomial $ P(z)=4z^{5}+2z^{4}-20z^{3}-5z^{2}+11z-2$. Thanks ^_^ Also I keep seeing it around how do you make the words blue and one needs to click on the words to make stuff appear? The link thing.

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Note that (1) $ (x - y)z = - (y - z)$, (2) $ (y - z)x = - (z - x)$, and (3) $ (z - x)y = - (x - y)$. Case 1: $ x$, $ y$, and $ z$ are distinct. Hence, $ xyz = - 1$. Therefore, \[ xy + y^2 = \frac {y}{2} + 1\,, \] \[ yz + z^2 = \frac {z}{2} + 1\,, \] and \[ zx + x^2 = \frac {x}{2} + 1\,. \] Call the three equations above (*). Thus, if $ s = x + y + z$, then \[ s^2 - xy - yz - zx = \frac {s}{2} + 3\,. \] However, \[ 2s + (xy + yz + zx) = (x + y + zx) + (y + z + xy) + (z + x + yz) = \frac {3}{2}\,. \] Therefore, \[ s^2 + 2s - \frac {3}{2} = \frac {s}{2} + 3\,. \] Consequently, $ s = - 3$ or $ s = \frac {3}{2}$. If $ x + y + z = s = - 3$, then $ xy + yz + zx = \frac {15}{2}\,.$ Note that \[ x^2 + y^2 + z^2 = s^2 - 2(xy + yz + zx) = - 6\,. \] Thus, $ x$, $ y$, or $ z$ is not real. Therefore, $ x + y + z$ must be $ \frac {3}{2}$. In this case, $ xy + yz + zx = - \frac {3}{2}$. Therefore, $ x,y,z$ are the roots of \[ t^3 - \frac {3}{2}t^2 - \frac {3}{2}t + 1 = 0\,. \] Hence, $ (x,y,z)$ is a permutation of $ \left(2, - 1,\frac {1}{2}\right)$. *PS: Alternatively, if $ s = - 3$, from (*), \[ (z - x)y = \left(y + z - \frac {1}{2}\right)(y - z) = - \left(x + \frac {7}{2}\right)(y - z)\,, \] \[ (x - y)z = \left(z + x - \frac {1}{2}\right)(z - x) = - \left(y + \frac {7}{2}\right)(z - x)\,, \] and \[ (y - z)x = \left(x + y - \frac {1}{2}\right)(x - y) = - \left(z + \frac {7}{2}\right)(x - y)\,. \] Hence, \[ - 1 = xyz = - \left(x + \frac {7}{2}\right)\left(y + \frac {7}{2}\right)\left(z + \frac {7}{2}\right)\,. \] Thus, \[ 1 = \frac {343}{8} + \frac {49}{4}(x + y + z) + \frac {7}{2}(xy + yz + zx) + xyz\,, \] or $ 1 = \frac {343}{8} + \frac {49}{4}( - 3) + \frac {7}{2}\left(\frac {15}{2}\right) - 1 = \frac {251}{8}$, which is a contradiction. Thus, $ s = \frac {3}{2}$. *********** Case 2: Two of $ x$, $ y$, and $ z$ are equal. Then, (1), (2), and (3) ensure that $ x = y = z$. In this case, they are the roots of \[ t^2 + 2t - \frac {1}{2} = 0\,. \] That is, $ x = y = z = - 1\pm\sqrt {\frac {3}{2}}\,.$

Each equation is equivalent with $ (1+x)(1+y)=\frac{3}{2}$ $ (1+y)(1+z)=\frac{3}{2}$ $ (1+z)(1+x)=\frac{3}{2}$ It's obvious that $ (1+x)(1+y)(1+z)=\frac{3}{2} \sqrt{{\frac{3}{2}}}$ or $ (1+x)(1+y)(1+z)=-\frac{3}{2} \sqrt{\frac{3}{2}}$ Because $ x,y,z \ne -1$, we can have $ 1+z = \sqrt{\frac{3}{2}}$ $ 1+x= \sqrt{\frac{3}{2}}$ $ 1+y=\sqrt{\frac{3}{2}}$ or $ 1+z = -\sqrt{\frac{3}{2}}$ $ 1+x= -\sqrt{\frac{3}{2}}$ $ 1+y=-\sqrt{\frac{3}{2}}$ Hence, the solutions are $ (\sqrt{\frac{3}{2}}-1,\sqrt{\frac{3}{2}}-1,\sqrt{\frac{3}{2}}-1)$ and $ -\sqrt{\frac{3}{2}}-1, -\sqrt{\frac{3}{2}}-1, -\sqrt{\frac{3}{2}}-1)$.