Solve for positive real solutions of cyclic equations $x+y^2+2xy=9$, $y+z^2+2yz=47$, $z+x^2+2xz=16$

Question

Solve over positive reals $$x+y^2+2xy=9$$ $$y+z^2+2yz=47$$ $$z+x^2+2xz=16$$

With standard manipulation we get that $x+y+z=8$. Thus $x=8-y-z$ and we have two equations, $$(8-y-z)^2+y^2+2(8-y-z)y=9$$ $$y+z^2+2yz=47$$ Which becomes $$-y^2-2yz+15y-z=1$$ $$y+z^2+2yz=47$$ Adding up gets $$-y^2+16y+z^2-z=1$$ $$\implies z^2=z+1+y^2-16y$$ And this substitutes to $$y+z+1+y^2-16y+2yz=47$$ $$\implies y+\bigg(\frac{-y^2+15y-1}{2y+1}\bigg)+1+y^2-16y+2y\bigg(\frac{-y^2+15y-1}{2y+1}\bigg)=47$$ However, this results into a nasty quartic which I don't know how to simplify. Thanks!

Answer 1

Continue by substituting $z=\frac{-y^2+15y-1}{2y+1}$ into $y+z^2+2yz=47$,

$$y+\left(\frac{-y^2+15y-1}{2y+1}\right)^2+2y\left(\frac{-y^2+15y-1}{2y+1}\right)=47$$

Simplify,

$$-3y^4+32y^3+69y^2-219y-46=0$$

and factorize

$$(y-2)(3y^3-26y^2-121y-23)=0$$

which yields the positive real solutions $x=1,\>y=2,\>z=5$. The other three sets of solutions are also real, but containing negative values.

Answer 2

$\begin{cases} x+y^2+2xy=9\\ y+z^2+2yz=47\\ z+x^2+2xz=16 \end{cases} \implies \begin{cases} (2 x + 2 y)^2 - (2 x - 1)^2 = 35\\ (2 y + 2 z)^2 - (2 y - 1)^2 = 187\\ (2 z + 2 x)^2 - (2 z - 1)^2 = 63 \end{cases} \implies \begin{cases} x=1\\ y=2\\ z=5 \end{cases}$