Solution of the given system of equation

Question

For $n\geq3$, determine all real solutions of the system of $n$ equations. $$x_{1}+x_{2}+\cdots+x_{n-1}=\frac{1}{x_{n}}$$ $$\vdots$$ $$x_{1}+\cdots+x_{i-1}+x_{i+1}+\cdots+x_{n}=\frac{1}{x_{i}}$$ $$\vdots$$ $$x_{2}+\cdots+x_{n}=\frac{1}{x_{1}}$$ Note: I have added $x_{i}$ to both the side of the equation to get $x_{i}+\frac{1}{x_{i}}=s$ for some real constant $s$. But after solving the quadratic equation I was unable to solve the problem further.

Answer 1

Hint:

Denote $s = \sum_{i=1}^nx_i$, then $$s =x_1+\frac{1}{x_1}=...=x_n+\frac{1}{x_n}$$ The $(x_i)_i$ are solutions of the quadratic equation $$Z^2-sZ+1=0$$ or for $i=1,...,n$ $$\color{red}{x_i\in \left\{\frac{s+\sqrt{s^2-4}}{2},\frac{s-\sqrt{s^2-4}}{2}\right\}=: \left\{z_1,z_2\right\}} \tag{1}$$ Suppose there are $p\in \{0,1,...,n \}$ values of $z_1$ (then $n-p$ values of $z_2$), then $$s = pz_1+(n-p)z_2$$ $$\begin{align} &\implies (s-nz_2)=p(z_1-z_2) = p\sqrt{s^2-4}\\ &\implies \left(s-n\frac{s-\sqrt{s^2-4}}{2}\right) = p\sqrt{s^2-4}\\ &\implies \left(\frac{n}{2}-p\right)s = \left(\frac{n}{2}-1\right)\sqrt{s^2-4} \tag{2} \end{align}$$

• If $p \le n/2$: $s$ must be negative, then $$(2)\implies \left(\frac{n}{2}-p\right)^2 s^2 = \left(\frac{n}{2}-1\right)^2(s^2-4) $$ $$\implies s^2 = \frac{4\left(\frac{n}{2}-1\right)^2}{\left(\frac{n}{2}-1\right)^2 -\left(\frac{n}{2}-p\right)^2}$$ $$\implies \color{red}{s = -\frac{2\left(\frac{n}{2}-1\right)}{\sqrt{\left(\frac{n}{2}-1\right)^2 -\left(\frac{n}{2}-p\right)^2}}} \tag{3}$$ (attention, here, we need to study the cases where $p = 0$ or $p = 1$)
• If $p > n/2$: $s$ must be positive, then $$(2)\implies s^2 = \frac{4\left(\frac{n}{2}-1\right)^2}{\left(\frac{n}{2}-1\right)^2 -\left(\frac{n}{2}-p\right)^2}$$ $$\implies \color{red}{s = \frac{2\left(\frac{n}{2}-1\right)}{\sqrt{\left(\frac{n}{2}-1\right)^2 -\left(\frac{n}{2}-p\right)^2}}} \tag{4}$$

I let the conclusion to you, it suffices to use the 3 results $(1), (3), (4)$.