Solve the system $x^2+y^2+z^2=2$ and $x+y+z = 2+xyz$, where $x,y,z$ are real numbers.

Question

Solve the system $x^2+y^2+z^2=2$ and $x+y+z = 2+xyz$, where $x,y,z$ are real numbers.

My idea was to write $$a=-x-y-z$$ $$b=xy+yz+zx$$ $$c=-xyz$$ then $x,y,z$ are solution of $$x^3+ax^2+bx+c=0$$ where $a^2-2b=2$ and $-a=2-c$ so $$2x^3+2ax^2+(a^2-2)x+2a+4=0$$ but could not solve this third degree equation.

Any idea?

Answer 1

Brute force: let $=+/3$ and write the equation in $$ as $^3+()+()=0$. The discriminant $−4^3()−27^2()$ is non-negative only when $=−2$. That forces one of the $$,$$,$$ to be 0 and all solutions are of the form $(0,1,1)$ or re-arrangements.

Answer 2

\begin{align} x^2+y^2+z^2&=2 \tag{1}\label{1} ,\\ x+y+z &= 2+xyz \tag{2}\label{2} ,\\ a&=-(x+y+z) \tag{3}\label{3} , \end{align}

\begin{align} 2x^3+2ax^2+(a^2-2)x+2a+4&=0 \tag{4}\label{4} \end{align}

The discriminant of \eqref{4} is \begin{align} \Delta&= 18\cdot 2\cdot (2\cdot a)\cdot (a^2-2)\cdot (2\cdot a+4)-4\cdot (2\cdot a)^3\cdot (2\cdot a+4) \\ &\phantom{=18}+(2\cdot a)^2\cdot (a^2-2)^2-4\cdot 2\cdot (a^2-2)^3-27\cdot 2^2\cdot (2\cdot a+4)^2 \\ &= -4(a^4-4a^3-16a^2+40a+104)(a+2)^2 \tag{5}\label{5} , \end{align}

$\Delta\ge0$, that is, all three roots are real, is true only for $a=-2$.

Hence, from \eqref{2}, we must have \begin{align} xyz&=0 \tag{6}\label{6} , \end{align}

so one and only one of the variables must be zero, and both of the other two must be $1$.