How to solve the following system of $4$ nonlinear equations

Question

How to solve this equation system?

$$\frac{-1}{x_1^{2} x_2 x_3} +y(x_2+2x_3) = 0, $$

$$x_3- \frac{1}{x_1 x_2^{2} x_3} + y x_1 = 0, $$

$$x_2 - \frac{1}{x_1 x_2 x_3^{2}} + 2yx_1 = 0, $$

$$x_1 x_2 + x_1 x_3 = 4 $$

where $x_1,x_2,x_3,y \neq 0$.

I have tried, but I can't get something. It seems unclear how to express one variable in terms of the others.

Answer 1

If you are satisfied with a computational solution then one may consider the variety $V \subset \mathbb{A}^4_{x_1, x_2, x_3, y}$ given by

\begin{align*} x_1^2 x_2^2 x_3 y + 2 x_1^2 x_2 x_3^2 y - 1 = 0 \\ x_1^2 x_2^2 x_3 y + x_1 x_2^2 x_3^2 - 1 =0\\ 2 x_1^2 x_2 x_3^2 y + x_1 x_2^2 x_3^2 - 1 =0\\ x_1 x_2 + x_1 x_3 - 4 =0 \end{align*} these are the numerators of your rational functions - i.e., this condition holds when $x_1,x_2,x_3,y \neq 0$ if and only if your condition holds.

The variety $V$ has dimension $0$ and (over $\mathbb{Q}$) has exactly one irreducible component. However it has no rational points i.e., your original equation has no solution in $\mathbb{Q}$.

However, let $\alpha \in \mathbb{C}$ be a root of $f(x) = x^4 - 4x^2 - 12x +4$. Then over $K = \mathbb{Q}(\alpha)$ (and indeed over $\mathbb{C}$) there are exactly $4$ solutions to the original system, namely \begin{align*} (x_1, x_2, x_3, y) = \left( \frac{4\alpha^2 - 8}{9}, \frac{\alpha^3 - 2\alpha - 12}{4}, \frac{\alpha^3 - 2\alpha - 12}{8}, \frac{9\alpha^3 + 3\alpha^2 - 36\alpha - 114}{64} \right) \end{align*} and its Galois conjugates (i.e., replace $\alpha$ by the other $3$ roots of $f(x)$).

All of these claims were checked with Magma as follows:

F := Rationals();  // Replace this with the field you care about.
A<x1,x2,x3,y> := AffineSpace(F,  4);

f1 := Numerator( (-1)/(x1^2*x2*x3) + y*(x2 + 2x3));
f2 := Numerator( x3 - 1/(x1*x2^2*x3) + y*x1);
f3 := Numerator( x2 - 1/(x1*x2*x3^2) + 2*y*x1);
f4 := x1*x2 + x1*x3 - 4;

C := Cluster(A, [f1,f2,f3,f4]);

P<a> := PolynomialRing(F);
K<a> := quo<P | (a^4 - 4*a^2 - 12*a+ 4)>;

Degree(C) eq 4;
Points(C, K);

Answer 2

The solutions are:

$y=\frac{3^{5/3}}{2^{16/3}}$,

$x_{1}=\frac{2^{11/3}}{3^{4/3}}$,

$x_{2}=\frac{3^{1/3}}{2^{2/3}}$,

$x_{3}=\frac{3^{1/3}}{2^{5/3}}$.

From the first three equations we write the value of the variable y as a function of $x_{1}$, $x_{2}$ and $x_{3}$:

$y=\frac{1}{x_{1}^2x_{2} x_{3}(x_{2}+2x_{3})}=\frac{1-x_{1} x_{2}^{2} x_{3}^{2}}{ x_{1}^2 x_{2}^{2} x_{3}}=\frac{1-x_{1} x_{2}^{2} x_{3}^{2}}{2 x_{1}^{2} x_{2} x_{3}^{2}}$,

From the last equality we derive:

$x_{2}=2x_{3}$.

Consequently

$y=\frac{1}{x_{1}^{2}(2x_{3}) x_{3}(2x_{3}+2x_{3}}=\frac{1}{8x_{1}^{2} x_{3}^{3}}$.

From the second equation we derive:

$x_{1}=\frac{1}{8x_{3}^{4}}$,

and finally from the fourth equation we get:

$\frac{1}{8x_{3}^{4}}(2x_{3}+x_{3})=4$,

$\frac{3}{8x_{3}^3}=4$,

$x_{3}=\frac{6^{1/3}}{4}$.

Backwards we can calculate the values of the other unknowns.