$$\begin{cases}
x^2 = yz + 1 \\
y^2 = xz + 2 \\
z^2 = xy + 4
\end{cases}
$$
How to solve above system of equations in real numbers? I have multiplied all the equations by 2 and added them, then got $(x – y)^2 + (y – z)^2 + (x – z)^2 = 14$, but it leads to nowhere.
Best Answer
Multiplying both sides of the first equation by $z$, of the second by $x$, and of the third by $y$, we get $$x^2z=z^2y+z, \quad y^2x=x^2z+2x,\quad z^2y=y^2x+4y.$$ Adding up and cancelling, we get
$$2x+4y+z=0.$$ Similarly, $$x^2y=y^2z+y, \quad y^2z=z^2x+2z, \quad z^2x=x^2y+4x,$$ giving
$$4x+y+2z=0.$$
To finish, use the linear equations to express $y$ and $z$ in terms of $x$, and substitute in $x^2=yz+1$.