<s>[INST] For $n\ge 3$ determine all real solutions of the system of $n$ equations. 
Question: For $n\ge 3$ determine all real solutions of the system of $n$ equations: $$x_1+x_2+\cdots+x_{n-1}=\frac{1}{x_n}\\ \cdots \\ x_1+x_2+\cdots+x_{i-1}+x_{i+1}+\cdots+x_n=\frac{1}{x_i}\\ \cdots \\ x_2+\cdots+x_{n-1}+x_n=\frac{1}{x_1}.$$ 

My approach: It is given that $$x_1+x_2+\cdots+x_{n-1}=\frac{1}{x_n}\\ \cdots \\ x_1+x_2+\cdots+x_{i-1}+x_{i+1}+\cdots+x_n=\frac{1}{x_i}\\ \cdots \\ x_2+\cdots+x_{n-1}+x_n=\frac{1}{x_1}.$$ 
Define $$S_n:=x_1+x_2+\cdots+x_n.$$
This implies that $$x_1+x_2+\cdots+x_{n-1}+x_n=x_n+\frac{1}{x_n}\\ \cdots \\ x_1+x_2+\cdots+x_{i-1}++x_i+x_{i+1}+\cdots+x_n=x_i+\frac{1}{x_i}\\ \cdots \\ x_1+x_2+\cdots+x_{n-1}+x_n=x_1+\frac{1}{x_1}.$$
Therefore, we have $$x_j+\frac{1}{x_j}=S_n, \forall 1\le j\le n.$$ 
Now for any $1\le j\le n,$ we have $$x_j+\frac{1}{x_j}=x_n+\frac{1}{x_n}\\ \implies \frac{1}{x_j}-\frac{1}{x_n}=x_n-x_j\\ \implies \frac{x_n-x_j}{x_nx_j}=x_n-x_j\\ \implies (x_n-x_j)\left(\frac{1}{x_nx_j}-1\right)=0\\ \implies x_j=x_n \text{ or } x_j=\frac{1}{x_n}.$$ 
Now let us have $i (0\le i\le n-1)$ of the $(n-1)$ numbers $x_j, 1\le j\le n-1$ such that $$x_j=x_n$$ and the rest $(n-1-i)$ numbers such that $$x_j=\frac{1}{x_n}.$$
Therefore, since $$x_1+x_2+\cdots+x_{n-1}=\frac{1}{x_n}\\\implies i.x_n+(n-1-i).\frac{1}{x_n}=\frac{1}{x_n}\\ \implies i.x_n^2+(n-2-i)=0\\\implies i.x_n^2=2+i-n.$$
Now since $x_n\in\mathbb{R}$ and $x_n\neq 0\implies x_n^2>0.$ Now since $i\ge 0\implies i.x_n^2\ge 0 \implies 2+i-n\ge 0\implies i\ge n-2.$ Therefore $i=n-2,n-1$. 
Now when $i=n-2$, we have $(n-2)x_n^2=0.$ Now since it is given that $n\ge 3\implies n-2\ge 1>0$. Therefore $x_n^2=0\implies x_n=0$. But $x_n\neq 0$. Therefore $i\neq n-2$. 
Now when $i=n-1$, we have $(n-1)x_n^2=1\implies x_n^2=\frac{1}{n-1}\implies x_n=\pm\frac{1}{\sqrt{n-1}}.$ Therefore we have $$x_1=x_2=\cdots=x_n=\pm\frac{1}{\sqrt{n-1}}.$$ 
Therefore the required set of solutions are $$(x_1,x_2,\cdots,x_n)=\left(\frac{1}{\sqrt{n-1}},\frac{1}{\sqrt{n-1}},\cdots,\frac{1}{\sqrt{n-1}}\right),\left(-\frac{1}{\sqrt{n-1}},-\frac{1}{\sqrt{n-1}},\cdots,-\frac{1}{\sqrt{n-1}}\right).$$
Can someone check if my solution is correct or not? And if correct, is there a more better and efficient solution than this?
 [/INST] We have $$x_1+x_2+\cdots+x_{n-1}-\frac{1}{x_n}=0\\x_1+x_2+\cdots-\frac{1}{x_{n-1}}+x_n=0\\.........................\\.........................\\-\frac{1}{x_1}+x_2+\cdots+x_{n-1}+x_n=0 $$ It follows
$$\sum_{i=1}^{i=n}\left((n-1)x_i-\frac{1}{x_i}\right)=\sum_{i=1}^{i=n}\frac{(n-1)x_i^2-1}{x_i}=0 $$
Then a clear solution is given by $(n-1)x_i^2-1=0$ for all $i$. 
Thus the two solutions given by the O.P.
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