Matematické Fórum

stuart clark · 20. 03. 2012 13:52

Calculate $x,y,z\in \mathbb{R}$ in system of equations

$x^2=1+2xy$

$y^2=1+2yz$

$z^2=1+2zx$

laszky · 12. 03. 2018 03:57 — Editoval laszky (13. 03. 2018 21:32)

System has 6 solutions $(x,y,z)\in\{(p,q,r), (-p,-q,-r), (q,r,p), (-q,-r,-p), (r,p,q), (-r,-p,-q)\}$ , where $p^2,q^2,r^2$ are three distinct (real and positive) solutions of the cubic equation $7x^3-35x^2+21x-1=0$ . They can be expressed in the form

$x_k = \frac{8}{3}\cos\left(\frac{1}{3}\mathrm{arccos}\left(\frac{13}{14}\right)+\frac{2k\pi}{3}\right) + \frac{5}{3},\quad k=0,1,2$ .

We choose $p=\sqrt{x_0}, q=-\sqrt{x_2}, r=\sqrt{x_1}$

Edit: The polynomial can be obtained by expressing $x=x(z)=\frac{z^2-1}{2z}, y=y(x)=\frac{x^2-1}{2x}, z=z(y)=\frac{y^2-1}{2y}$ and then evaluating $x = x(z(y(x)))$ .

vanok · 13. 03. 2018 20:23

Hi ↑ laszky:,
Please, you could develop method to arrive towards $7x^3-35x^2+21x-1=0$ .
Thank you.

Matematické Fórum

#1 20. 03. 2012 13:52

system of equations

#2 12. 03. 2018 03:57 — Editoval laszky (13. 03. 2018 21:32)

Re: system of equations

#3 13. 03. 2018 20:23

Re: system of equations

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