Solve three-variable system $x^2 - yz = 1,\>y^2 - xz = 2,\>z^2 - xy = 3$

Question

I am going around in circles on this system

$$x^2 - yz = 1\\ y^2 - xz = 2\\ z^2 - xy = 3$$

I have tried a few things (below) but keep hitting a wall. I know that by adding the three equations we get: $x^2 + y^2 + z^2 - yz - xz - xy =6$. And also that:

$(x-y)^2 + (x-z)^2 + (y-z)^2 = 2x^2+2y^2+2z^2-2xy-2xz-2yz=12$

I have tried the following:
$(x-y)(x+y+z)=(x^2-yz)-(y^2-xz)=-1$
$(x-z)(x+y+z)=(x^2-yz)-(z^2-xy)=-2$
$(y-z)(x+y+z)=(y^2-xz)-(z^2-xy)=-1$

So I think it follows that $x-y=y-z$ and therefore $x-z=2(x-y)$

Substituting in:
$(x-y)^2+[2(x-y)]^2+(x-y)^2=12$
$6(x-y)^2=12$
$x-y=\sqrt{2}$
Therefore: $y-z=\sqrt{2}$ and $x-z=2\sqrt{2}$

But when I rearrange and substitute into one of the original equations, the solutions I get for x, y and z don't actually work.

$y=x-\sqrt{2}$ and $z=x-2\sqrt{2}$
Substituting into the first equation:
$x^2-(x-\sqrt{2})(x-2\sqrt{2})=1$
$x=\frac{3\sqrt{2}}{2}$
Giving $y=\frac{\sqrt{2}}{2}, z=-\frac{\sqrt{2}}{2}$

Unfortunately, substituting these values into the three original equations only satisfies the second one. Can anyone please help?

Answer 1

You're on the right track but made an algebra error (cf. $y - z$). These are quadratic equations, so there are two solutions:

$$\left\{x\to -\frac{5}{3 \sqrt{2}},y\to \frac{1}{3 \sqrt{2}},z\to \frac{7}{3 \sqrt{2}}\right\}$$

$$\left\{x\to \frac{5}{3 \sqrt{2}},y\to -\frac{1}{3 \sqrt{2}},z\to -\frac{7}{3 \sqrt{2}}\right\}$$

Answer 2

This type of the system is well known. Its general form reads

$$x^2 - yz = a,\>\>\>y^2 - xz = b,\>\>\>z^2 - xy = c$$

and has the solutions given in the symmetric expression below

$$(x,y,z)=\pm \frac{(a^2-bc, b^2-ca, c^2 -ab)}{\sqrt{a^3+b^3+c^3-3abc}} $$

So, substitute $a=1,\>b=2,\>c=3$ to obtain $\pm(\frac{-5}{3 \sqrt{2}},\frac{1}{3 \sqrt{2}},\frac{7}{3 \sqrt{2}})$.