Solving system of three quadratic equations

Question

$$\begin{cases} x^2 = yz + 1 \\ y^2 = xz + 2 \\ z^2 = xy + 4 \end{cases} $$

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.

Answer 1

Now I would note that $(\pm 3)^2+(\pm 2)^2+(\pm 1)^2=14$. Since $(x-y)+(y-z)=(x-z)$ we can't assign the signs and values arbitrarily. For example, from $x-y=1, y-z=2, x-z=3$ I get $x=1,y=0,z=-2$ and from $x-y=-1, y-z=-2, x-z=-3$ I get $x=-1,y=0,z=2$ The other possibilities do not give solutions.

Answer 2

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 $x^2=yz+1$.

Answer 3

Maple indicates that a Groebner basis for the corresponding polynomials is $[z^2 - 4, y, z + 2 x]$. Thus there are two solutions: $x = \pm 1, y = 0, z = \mp 2$.