Mathematical Reflections 2011, Issue 6 - Problem J212

Problem:
Solve in real numbers the system of equations
$$\begin{array}{lll} (x-2y)(x-4z)=6 \\ (y-2z)(y-4x)=10 \\ (z-2x)(z-4y)=-16. \end{array}$$

Proposed by Titu Andreescu.

Solution:

Rewriting the equations, we have
$$\begin{array}{lll} x^2-2xy+8yz-4zx=6 \\ y^2-4xy-2yz+8zx=10 \\ z^2+8xy-4yz-2zx=-16. \end{array}$$
and summing up the equations we obtain $(x+y+z)^2=0$, i.e. $x+y+z=0$. Putting $z=-x-y$, and substituting into the first two equations, we have
$$\begin{array}{lll} 5x^2-6xy-8y^2=6 \\ 3y^2-10xy-8x^2=10,\end{array}$$ so $5(5x^2-6xy-8y^2)-3(3y^2-10xy-8x^2)=49(x^2-y^2)=0$, which gives $y=\pm x$. If $y=x$, the first equation yields $-9x^2=6$, so no real solutions. If $y=-x$, the first equations yields $3x^2=6$, so $x=\pm \sqrt{2}$ and $y=\mp \sqrt{2}$. Therefore, the real solutions are $(\sqrt{2},-\sqrt{2},0)$ and $(-\sqrt{2},\sqrt{2},0)$.