Determine all real solutions of the system of n equations

Question

For $n\geq3$, determine all real solutions of the system of $n$ equations:
$x_{1}+x_{2}+...+x_{n-1}=\frac{1}{x_{n}}$
...
$x_{1}+x_{2}+...+x_{i-1}+x_{i+1}+...x_{n}=\frac{1}{x_{i}}$
...
$x_{2}+x_{3}+...+x_{n-1}+x_{n}=\frac{1}{x_{1}}$

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I added all the equations to get $(n-1)(x_{1}+x_{2}+x_{3}+...+x_{n-1}+x_{n})=\frac{1}{x_2}+\frac{1}{x_2}...+\frac{1}{x_{n-1}}+\frac{1}{x_n}$. I don't know what to do about it. I don't even know if adding the equations helps or not. No clue how to proceed.
Although by symmetry, I guess all the $x_i$'s are equal. Then the solution is $x_i=\frac{1}{\sqrt{n-1}},\frac{-1}{\sqrt{n-1}}$
Source:Test of Mathematics at the 10+2 level, Indian Statistical Institute

Answer 1

For this problem, you have to explain why "all variables equal" must be true. It is not immediately obvious.

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Subtracting equation $i$ and $j$, we get $ x_i - x_j = \frac{1}{x_j} - \frac{1}{x_i} \Rightarrow (x_i - x_j)(x_ix_j-1) = 0$

So, we either have $x_i = x_j$ or $x_i = \frac{1}{x_j}$.

Let $2 \leq k \leq n - 2$ of them be equal to $x$, and the rest equal to $\frac{1}{x}$. Then, we have the equation $ (k-1) x + (n-k-1) \frac{1}{x} = 0$, for which there are no real solutions. If they allow for complex solutions, we have $ x = \pm \sqrt{\frac{ n-k-1} { k-1} } i $. (Note that the fraction has positive terms, and is well defined.)

Let $k = n$ of them be equal to $x$. Then we have $(n-1) x = \frac{1}{x}$, or that $x = \pm \sqrt{ \frac {1}{n-1}} $ (which was what you found).

Let $k = n-1$ of them be equal to $x$. Then we have $(n-2) x = 0$, which yields the solution $x=0$ (since $n\geq 3$), and we reject this case.