How to prove the equations have only one real solution?

Question

There are $n$ equations.

I need answer for the case $n=3$.

$$ \frac{1}{x_1}(1-x_1)^2+\frac{1}{x_2}(1-x_2)^2+\cdots+\frac{1}{x_n}(1-x_n)^2=0, $$ and $$ \frac{1}{x_1}(1-x_1^2)^j+\frac{1}{x_2}(1-x_2^2)^j+\cdots+\frac{1}{x_n}(1-x_n^2)^j=0,\; 2\leq j\leq n,$$

where $x_j\geq 1$ or $x_j \leq -1$.

If $n=2,3,4$, computer told me that the only real solution is $(1,1,\cdots,1)$. I did not tell computer the condition $|x_j|\geq 1$.

I do not know if some inequality can be applied like chebyshev's inequality and rearrangement inequality. It seems also a little like the Vandermonde determinant.

Comments: : The following three answers perfectly solve the case $n=3$, two of them drop the condition $|x_j|\geq 1$. The case $n=3$ is enough for me, however, for more interested you can try with larger $n\geq 4$. I am also very curious about that.

Answer 1

I need answer for the case $n=3$.

We rename variables and obtain the following system of equations.

$$\cases{ \frac{(1-x)^2}{x}+\frac{(1-y)^2}{y}+\frac{(1-z)^2}{z} & =$\,$ 0\\ \frac{(1-x^2)^2}{x}+\frac{(1-y^2)^2}{y}+\frac{(1-z^2)^2}{z} & = $\,$0\\ \frac{(1-x^2)^3}{x}+\frac{(1-y^2)^3}{y}+\frac{(1-z^2)^3}{z} & = $\,$0} $$

Suppose for a contradiction that $(x,y,z)\ne (1,1,1)$. Then the system

$$\cases{ u+v+w & =$\,$ 0\\ (1+x)^2u+(1+y)^2v+(1+z)^2w & = $\,$0\\ (1+x)^2(1-x^2)u+(1+y)^2(1-y^2)v+(1+z)^2(1-z^2)w& = $\,$0\\} $$

of linear equations has a non-zero solution $(u,v,w)=\left(\frac{(1-x)^2}{x},\frac{(1-y)^2}{y},\frac{(1-z)^2}{z} \right)$.

Thus the system is degenerated, that is $$ 0=\left|\begin{matrix} 1 & 1 & 1\\ (1+x)^2 & (1+y)^2 & (1+z)^2\\ (1+x)^2(1-x^2) & (1+y)^2(1-y^2) & (1+z)^2(1-z^2) \end{matrix}\right|=$$ $$(x-y)(x-z)(y-z)((x^2y+x^2z+xy^2+xz^2+y^2z+yz^2)+2xyz+$$ $$2(x^2+y^2+z^2)+4(xy+xz+yz)+4(x+y+z)+2)=$$ $$(x-y)(x-z)(y-z)((x^2y+x^2z+xy^2+xz^2+y^2z+yz^2)+2xyz+$$ $$2(x+y+z)^2+4(x+y+z)+2).$$

Let $\sigma_1=x+y+z$, $\sigma_2=xy+xz+yz$, and $\sigma_3=xyz$ be the symmetric polynomials on variables $x$, $y$, and $z$. Then $x^2y+x^2z+xy^2+xz^2+y^2z+yz^2=\sigma_1\sigma_2-3\sigma_3$, $x^3+y^3+z^3=\sigma_1^3-3(\sigma_1\sigma_2-3\sigma_3)-6\sigma_3=$ $\sigma_1^3-3\sigma_1\sigma_2+3\sigma_3$, and $\frac 1x+\frac 1y+\frac 1z=\frac{\sigma_2}{\sigma_3}.$

The first two equations of the initial system imply $\frac{\sigma_2}{\sigma_3}-6+\sigma_1=0$ and $$\frac{\sigma_2}{\sigma_3}-2\sigma_1+\sigma_1^3-3\sigma_1\sigma_2+3\sigma_3=0.$$ It follows $$3\sigma_1\sigma_2-3\sigma_3=\frac{\sigma_2}{\sigma_3}-2\sigma_1+\sigma_1^3=6-\sigma_1-2\sigma_1+\sigma_1^3.$$

Suppose first that $x$, $y$, and $z$ are pairwise distinct. Then $$0=(x^2y+x^2z+xy^2+xz^2+y^2z+yz^2)+2xyz+2(x+y+z)^2+4(x+y+z)+2=$$ $$\sigma_1\sigma_2-3\sigma_3+2\sigma_3+2\sigma_1^2+4\sigma_1+2=$$ $$\sigma_1\sigma_2-\sigma_3+2\sigma_1^2+4\sigma_1+2=$$ $$2-\sigma_1+\frac 13\sigma_1^3+2\sigma_1^2+4\sigma_1+2=$$ $$\frac 13\sigma_1^3+2\sigma_1^2+3\sigma_1+4.$$

The only real root of the latter equation is $\sigma_1=-A-A^{-1}-2\approx -4.613$, where $A=\sqrt[3]{5+2\sqrt{6}}$. We have $\sigma_2=(\sigma_1-6)\sigma_3$ and so $$3\sigma_1(\sigma_1-6)\sigma_3-3\sigma_3=3\sigma_1\sigma_2-3\sigma_3=6-3\sigma_1+\sigma_1^3.$$ Thus $$\sigma_3=\frac{6-3\sigma_1+\sigma_1^3}{3\sigma_1(\sigma_1-6)-3}\mbox{ and }\sigma_2=(\sigma_1-6)\frac{6-3\sigma_1+\sigma_1^3}{3\sigma_1(\sigma_1-6)-3}.$$

Put $p(t)=t^3-\sigma_1t^2+\sigma_2t-\sigma_3$. By Vieta's formulae, $x$, $y$, and $z$ are roots of the equation $p(t)=0$. But the discriminant of the polynomial $p(t)$ equals $$\sigma_1^2\sigma_2^2-4\sigma_2^3-4\sigma_1^3\sigma_3-27\sigma_3^2+18\sigma_1\sigma_2\sigma_3\approx -21.702<0,$$ so the polynomial $p(t)$ has only one real root, a contradiction.

Suppose now that $x$, $y$, and $z$ are not pairwise distinct. Permuting $x$, $y$, and $z$, if needed, we can assume that $y=z$. Then $$\cases{ \frac{(1-x)^2}{x}+2\frac{(1-y)^2}{y}& =$\,$ 0\\ \frac{(1-x^2)^2}{x}+2\frac{(1-y^2)^2}{y} & = $\,$0\\ \frac{(1-x^2)^3}{x}+2\frac{(1-y^2)^3}{y} & = $\,$0\\ }.$$

From the first and the second equation we obtain $$(1+x)^2\cdot (-2)\frac{(1-y)^2}{y}+2\frac{(1-y^2)^2}{y}=0$$ $$-(1+x)^2+(1+y)^2=0$$ $$1+x=\pm (1+y)$$

If $x=y$ then $x=y=z$ and so $3\frac{(1-x)^2}{x}=0$ and $(x,y,z)=(1,1,1)$.

If $x\ne y$ then $y=-2-x$.

From the second and the third equation we obtain $$(1-x^2)\cdot (-2)\frac{(1-y^2)^2}{y}+2\frac{(1-y^2)^3}{y}=0.$$

Since $y\ne \pm 1$, we can divide the equality by $2\frac{(1-y^2)^2}{y}$ and obtain $-(1-x^2)+1-y^2=0$, so $x^2=y^2$. Since $x\ne y$, we have $y=-x=-2-x$, a contradiction.

Answer 2

Solution for $n=3$

For $|x|,|y|,|z|\ge1$, we aim to solve the system \begin{align}\frac{(1-x)^2}x+\frac{(1-y)^2}y+\frac{(1-z)^2}z&=0\\\frac{(1-x^2)^2}x+\frac{(1-y^2)^2}y+\frac{(1-z^2)^2}z&=0\\\frac{(1-x^2)^3}x+\frac{(1-y^2)^3}y+\frac{(1-z^2)^3}z&=0.\end{align} After expanding each equation and subtracting appropriately, we find an interesting equivalence \begin{align}x+y+z+\frac1x+\frac1y+\frac1z&=6\\x^3+y^3+z^3+3\left(\frac1x+\frac1y+\frac1z\right)&=12\\x^5+y^5+z^5+5\left(\frac1x+\frac1y+\frac1z\right)&=18\end{align} which we can attack using the elementary symmetric polynomials $e_1=\sum_{\rm cyc}x,e_2=\sum_{\rm cyc}xy,e_3=xyz$. Using Newton's identities, this is equivalent to \begin{align}e_1+\frac{e_2}{e_3}&=6\\e_1^3-3e_1e_2+3e_3+\frac{3e_2}{e_3}&=12\\e_1^5-5e_1^3e_2+5e_1^2e_3+5e_1e_2^2-5e_2e_3+\frac{5e_2}{e_3}&=18.\end{align}

The rest of my original solution which is quite tedious can be found here, and I will illustrate a much simpler solution by @MykolaPochekai:

Letting $e_2=re_3$, we have $e_1=6-r$ and $e_3=\dfrac{(12-3r)-(6-r)^3}{3(r^2-6r+1)}$. Substituting into the third equation yields $$\frac{(r-3)^4(r^3-24r^2+189r-498)}{9(r^2-6r+1)}=0.$$ Note that $|r|\le3$ from $|x|,|y|,|z|\ge1$, and it is easy to show the cubic has no real roots over this interval. Therefore, $e_1=3$, $e_2=3$ and $e_3=1$ and Vieta tell us each of $x,y,z$ solves $t^3-3t^2+3t-1=0$ so $x=y=z=1$ is the unique solution. Note that for completeness, the denominator is zero when $r=3-2\sqrt2$ but this is irrelevant since $12-3r-(6-r)^3\ne0$ and $e_3$ is finite. $\square$

Answer 3

Let $x_1+x_2+x_3=3u$, $x_1x_2+x_1x_3+x_2x_3=3v^2,$ where $v^2$ may be negative and $x_1x_2x_3=w^3$.

Thus, $$\sum_{cyc}\frac{(x_1-1)^2}{x_1}=0$$ gives $$u-2+\frac{v^2}{w^3}=0,$$ $$\sum_{cyc}\frac{(x_1^2-1)^2}{x_1}=0$$ gives $$9u^3-9uv^2+w^3-2u+2-u=0$$ and we obtain: $$v^2=\frac{(9u^3-3u+2)(2-u)}{18u-9u^2-1}$$ and $$w^3=\frac{9u^3-3u+2}{18u-9u^2-1}.$$ About the case $u=2$ we'll say later.

Also, $$\sum_{cyc}\frac{(x_1^2-1)^3}{x_1}=0$$ gives $$81u^5-135u^3v^2+45uv^4+15u^2w^3-5v^2w^3-(27u^3-27uv^2+3w^3)+3u+u-2=0$$ and we obtain: $$(u-1)^4(9u^3+18u^2+9u+4)(18u-9u^2-1)=0.$$ If $u=2$, so $v^2=0$ and $w^3=-68$, but the equation $$x^3-6x^2+68=0$$ has two complex roots.

Thus, $u\neq2$ and from here $18u-9u^2-1\neq0.$

Also, since $$\prod_{cyc}(x_1-x_2)^2=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)\geq0,$$ we obtain: $$\frac{(u-1)^6(9u^3-27u-14)(9u^3-3u+2)}{(18u-9u^2-1)^3}\geq0.$$ Also, $u^2\geq v^2$ gives $$u^2\geq \frac{(9u^3-3u+2)(2-u)}{18u-9u^2-1}$$ or $$\frac{(u-1)^2}{18u-1-9u^2}\leq0,$$ which gives $18u-1-9u^2<0$ or $u=1$.

Now, let $u\neq1$.

Thus, $18u-1-9u^2<0$, $(9u^3-27u-14)(9u^3-3u+2)\leq0$ and $9u^3+18u^2+9u+4=0,$ which is impossible.

Thus, $u=1$, $v^2=w^3=1$, which gives $x_1=x_2=x_3=1.$

We see that the condition $x_j\geq1$ or $x_j\leq-1$ is not necessary.