# Solving equations system: $xy+yz=a^2,xz+xy=b^2,yz+zx=c^2$

Solve the following system of equations for $x,y,z$ as $a,b,c\in\Bbb{R}$

\begin{align*}xy+yz&=a^2\tag{1}\\xz+xy&=b^2\tag{2}\\yz+zx&=c^2\tag{3}\end{align*}

My try: Assume that $x,y,z\ne 0$ (it is easy to check the case where some are zero).

Subtract first equation from second equation we will get the system \begin{align*}xz-yz&=b^2-a^2\\xz+yz&=c^2\end{align*} Adding and subtract both equations we get $$x=\frac{b^2+c^2-a^2}{2z},y=\frac{a^2+c^2-b^2}{2z}$$I tried to set the expressions into one of the equations and get value of $z$, but it got quiet messy.

Is there any easier way of solving the system?

Any help will be appreciated, thanks!

Adding we get $$2(xy+yz+zx)=a^2+b^2+c^2$$

$$2xy=a^2+b^2+c^2-2c^2=a^2+b^2-c^2$$ etc.

Muliplying we get $$8x^2y^2z^2=\prod_{\text{cyc}}(a^2+b^2-c^2)$$

$$xyz=\pm\sqrt{\dfrac{\prod_{\text{cyc}}(a^2+b^2-c^2)}8}$$

and $xy=\dfrac{a^2+b^2-c^2}2$

$$z=\dfrac{xyz}{xy}=?$$

• I don't understand the notation $\displaystyle\prod_{\text{cyc}}$. Commented Jul 13, 2016 at 8:07
• @Galc127, $$\prod_{\text{cyc}}(a^2+b^2-c^2)$$ $$=(a^2+b^2-c^2)(b^2+c^2-a^2)(c^2+a^2-b^2)$$ Commented Jul 13, 2016 at 8:08

In M2

S=QQ[a,b,c]
FS=frac S
R=FS[x,y,z]
I=ideal(x*y+y*z-a^2,x*z+x*y-b^2,y*z+z*x-c^2)
gens gb I -- | y+(a2+b2-c2)/(a2-b2-c2)z x+(-a2-b2+c2)/(a2-b2+c2)z z2+(a4-2a2b2+b4-c4)/(2a2+2b2-2c2) |