Challenge: If we solve for all roots, real or complex, we would find there are four solutions. (a) Show that all four solutions are real. (b) Show by direct derivation that they are as follows: $(1,1,1)$, $(A,B,C)$, $(B,C,A)$, $(C,A,B)$ $A=1+2\cos\frac{10\pi}{63}+2\cos\frac{13\pi}{63}+2\cos\frac{46\pi}{63}+2\cos\frac{41\pi}{63}+\frac{4}{3}\cos\frac{4\pi}{9}+\frac{2}{3}\cos\frac{8\pi}{9}$ $B=1+2\cos\frac{50\pi}{63}+2\cos\frac{61\pi}{63}+2\cos\frac{22\pi}{63}+2\cos\frac{47\pi}{63}+\frac{4}{3}\cos\frac{2\pi}{9}+\frac{2}{3}\cos\frac{4\pi}{9}$ $C=1+2\cos\frac{2\pi}{63}+2\cos\frac{53\pi}{63}+2\cos\frac{16\pi}{63}+2\cos\frac{17\pi}{63}+\frac{4}{3}\cos\frac{8\pi}{9}+\frac{2}{3}\cos\frac{2\pi}{9}$

How obtain the trigonometic solutions?

Challenge (a) $\begin{cases}x^2 - y = (z -1)^2\\y^2 - z = (x -1)^2 \\z^2 - x = (y - 1)^2\end{cases}$ Let $~~x-1=X, ~~y-1=Y, ~~z-1=Z$ (motivation: to get rid of the independent term), the system becomes: $\Longrightarrow \begin{cases} X^2 +2X-Y=Z^2~~~~~~~~(1)\\Y^2+2Y-Z = X^2~~~~~~~~(2) \\Z^2+2Z-X = Y^2~~~~~~~~(3)\end{cases}$ $(1)+(2)+(3)\Longrightarrow X+Y+Z=0~~~~~~~(4)$ $\Longrightarrow (X,Y,Z)=(0,0,0)\Rightarrow (x,y,z)=(1,1,1)$ Case where $(X,Y,Z)\neq (0,0,0)$: $(1)+2*(3)\Longrightarrow X^2-Y+Z^2+4Z=2Y^2~~~~~~~(5)$ $(5)+(2)\Longrightarrow Z^2+3Z=Y^2-Y~~~~~~~~~~~~~~~~~~~~~~~(6)$ $(4)\Longrightarrow X^2=(Y+Z)^2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(7)$ $(2)+(7)\Longrightarrow (Y+Z)^2=Y^2+2Y-Z\Longrightarrow Y=\frac{Z^2+Z}{2(1-Z)}~~~~~~(8)$ $(6)+(8)\Longrightarrow \begin{cases}Z^2+3Z=Y^2-Y\\Y=\frac{Z^2+Z}{2(1-Z)}\end{cases}$ $4(Z+3)(1-Z)^2=Z^3+4Z^2+Z-2\Longrightarrow 3Z^3-21Z+14=0$ It can be seen that this cubic has 3 real roots: $(X_1,Y_1,Z_1)=(-2.931,~2.211,~0.72),~~(X_2,Y_2,Z_2)=(2.211,~0.72,~-2.931),~~(X_3,Y_3,Z_3)=(0.72,~-2.931,~2.211)$ $(x_1,y_1,z_1)=(-1.931,~3.211,~1.72),~~(x_2,y_2,z_2)=(3.211,~1.72,~-1.931),~~(x_3,y_3,z_3)=(01.72,~-1.931,~3.211)$ Challenge (b) Applying the formulae for solving a cubic by means of trigonometry, the "cos" apparently vanishes. Maybe it it not the right formulae: see here https://en.wikibooks.org/wiki/Trigonometry/The_solution_of_cubic_equations