simplify system of equations -- any relation with golden ratio?

Question

I have a system of equations:

$w(1-v) = v(1+u) \tag{1}$

$(w-1)(u+1) = v-u^2 \tag{2}$

$u^2 (v+1)(w+3)+2 u v (w+1)-v (w+3)+w-1=0 \tag{3}$

First, I would like to know if this system of equations could be further simplified (specially the third equation). You can assume that $u,v,w$ are real numbers with values close to $u=0.389, v=0.469, w=1.229$.

Second (bonus question): Are the solutions to this system related in any interesting way to the golden ratio $\phi$.

Edit: I added a figure below to give an idea where $u,v,w$ came from. In this figure, $u,v,w$ are the (absolute value of) inverse of the slope of the green, red and blue lines.

Answer 1

The roots $u,\,v,\,w = 0.389, 0.469, 1.229$ are roots of cubics with coefficients that are powers of the golden ratio $\phi$ namely,

$$u^3 - 3\phi^2 u^2 + \phi^6u - \phi^2\sqrt{5} = 0$$ $$v^3 + 3\phi^{-2}v^2 + \phi^{-6}v - \phi^{-2} = 0$$ $$w^3 + 3w^2 + \phi^{-2}w - \phi^4 = 0$$

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Edit: Courtesy of Somos' answer, we have,

$$u = \phi-w\\ v = w/\phi^2$$

I find $w$ can also be expressed as the nice cubic,

$$\frac{(w+1)^3}{w+3}=\phi^2$$

Using trigonometric or hyperbolic functions, we can solve it simply as,

$$w = -1+\frac{2\,\phi\,\cos\left(\tfrac13\,\cos^{-1}\left(\frac{3\sqrt3}{\phi}\right)\right)}{\sqrt3} \approx 1.229$$ $$w = -1+\frac{2\,\phi\,\cosh\left(\tfrac13\,\cosh^{-1}\left(\frac{3\sqrt3}{\phi}\right)\right)}{\sqrt3} \approx 1.229 $$

where $\cos^{-1}x=\text{arccos}\,x$. More traditionally,

$$w = -1+x_1^{1/3}+x_2^{1/3}\approx 1.229\tag{1.1}$$

and $x_i$ are the two roots of,

$$x^2-2\phi^2\,x+\frac{\phi^6}{27}=0$$

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$\color{blue}{\text{Further edit:}}$ To address a comment of the OP, the solution to the general cubic,

$$x^3+ax^2+bx+c=0$$

is,

$$3x = -a+z_1^{1/3}+z_2^{1/3}\tag1$$

where the $z_i$ are the two roots of the quadratic Lagrange resolvent,

$$z^2 + (2 a^3 - 9 a b + 27 c) z + (a^2 - 3 b)^3 = 0$$

Alternatively,

$$3x = -a+z_1^{1/3}+\frac{a^2-3b}{z_1^{1/3}}\tag2$$

where $z_1$ is any non-zero root of the resolvent.

The advantage of $(1)$ is it somehow more symmetrical (which I prefer), while the advantage of $(2)$ is you take the cube root of only one value (which is the default of Mathematica) and is useful when the $z_i$ are complex.

Answer 2

eliminating step by step the variables we get for $u$ the equation $$(u^6-9u^5+27u^4-26u^3+u^2-15u-5)(u+1)=0$$ by a numerical method we get $${u = 0.3892773008780011471324554158555669542330590528674325981347375567503\dots\\ v = 0.4693432908633176939543775508532342246571802975697442128254750629456 \dots\\ w = 1.2287566878718937010721314185100711634872501269383302640007110659549\dots}$$

Answer 3

The three equations can be written as polynomials in $u,v,w$ as: $p_1 := w (1-v)-v (1+u)=0,\quad p_2 := (w-1) (u+1)-v+u^2=0,$ $p_3:= u^2 (v+1) (w+3)+2 u v (w+1)-v (w+3)+w-1=0.$

We use resultants where $\textrm{res}_x(p(x),q(x))$ is the resultant of polynomials in variable $x$. Compute $r_1 := \textrm{res}_v(p_1,p_2)/(u+1),\quad$ $r_2 := \textrm{res}_v(p_1,p_3)/(u+1),\quad$ $r_3 := \textrm{res}_u(p_1,p_2)/(v-1),\quad$ $r_4 := \textrm{res}_u(p_1,p_3)/(v-1),\quad$ $p_u := \textrm{res}_w(r_1,r_2) = u^6 -9u^5 +27u^4 -26u^3 +u^2 +15u -5,$ $p_v := -\textrm{res}_w(r_3,r_4)/v^4 = v^6 +9v^5 +27v^4 +18v^3 -5v^2 -7v +1,\quad$ $p_w =\textrm{res}_u(r_1,r_2) = w^6 +6w^5 +12w^4 +2w^3 -20 w^2 -18w +1.$

$p_u,p_v,p_w\;$ are irreducible. Note that $\;u = \phi-w,\;\;v = w/\phi^2,\;$ and $\;u v - w (u-v+w-1) = 0.$

Note that $u=-1,v=1$ is a solution to all three of the original equations.

Edit: There is still the mystery of equation $(3)$. How to get it given the numerical values of $\;u,v,w.$ That is, using the PARI/GP function $\texttt{lindep(v)}$ I found linear relations between $(u,v,w,\phi)$ and I can find polynomial equations like $p_1$ and $p_2,$ but they are satisfied if $u=\phi-w,\; v=w/\phi^2,\;$ and so far I can't get another integer polynomial in $\;u,v,w\;$ that I can solve for $\;w.$