# 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?

I get the same result. Here's my work.

I abhore "..."s, so I'll write the equations like this:

For $$i=1$$ to $$n$$,

$$\sum_{k=1, k\ne i}^n x_k =\dfrac1{x_i}$$.

Filling in the missing term,

$$\sum_{k=1}^n x_k =\dfrac1{x_i}+x_i$$.

Letting $$S = \sum_{k=1}^n x_k$$, we have $$S = \dfrac1{x_i}+x_i$$ so $$x_i^2-Sx_i+1 = 0$$ or $$x_i = \dfrac{S\pm\sqrt{S^2-4}}{2}$$.

Also, for $$i \ne j$$, $$\dfrac1{x_i}+x_i = \dfrac1{x_j}+x_j$$ or, multiplying by $$x_ix_j$$, $$x_j-x_i^2x_j =x_i-x_ix_j^2$$ or $$x_i-x_j =x_ix_j^2-x_i^2x_j =x_ix_j(x_j-x_i)$$.

If $$x_i \ne x_j$$ then $$x_ix_j = -1$$ so $$x_j = -\dfrac1{x_i}$$.

Set $$i = n$$ so the other values are either $$x_n$$ or $$-\dfrac1{x_n}$$.

Suppose $$m$$ of them are $$x_n$$. Then $$S =mx_n-(n-m)\dfrac1{x_n}$$ so $$x_n+\dfrac1{x_n} =mx_n-(n-m)\dfrac1{x_n}$$ or, writing $$x$$ for $$x_n$$, $$(m-1)x=(n-m+1)\dfrac1{x}$$.

We can't have $$m=1$$ for then $$\dfrac1{x} = 0$$.

Therefore $$m \ge 2$$ so that $$(m-1)x^2=(n-m+1)$$ or $$x =\pm\sqrt{\dfrac{n-m+1}{m-1}}$$.

I will choose $$x =\sqrt{\dfrac{n-m+1}{m-1}}$$ for now.

Then

$$\begin{array}\\ S &=m\sqrt{\dfrac{n-m+1}{m-1}}-(m-n)\sqrt{\dfrac{m-1}{n-m+1}}\\ &=x+\dfrac1{x}\\ &=\sqrt{\dfrac{n-m+1}{m-1}}+\sqrt{\dfrac{m-1}{n-m+1}}\\ \end{array}$$

so, multiplying by $$\sqrt{(n-m+1)(m-1)}$$,

$$m(n-m+1)-(m-n)(m-1) =(n-m+1)+(m-1)$$ or $$n =mn-m^2+m-(m^2-(n+1)m+n) =2m(n+1)-2m^2-n$$ or $$0 =2(m(n+1)-m^2-n)$$ or $$0 = m^2-(n+1)m+n$$ or $$0 =(m-1)(m-n)$$.

Since $$m > 1$$, we have $$m = n$$ so all the $$x_i$$ are the same, so $$S = \dfrac{n}{\sqrt{n-1}}$$ and each $$x_i =\dfrac1{\sqrt{n-1}}$$.

Note that $$\dfrac1{\sqrt{n-1}} +\sqrt{n-1} =\dfrac{1+n-1}{\sqrt{n-1}} =\dfrac{n}{\sqrt{n-1}} = S$$.

If $$x =-\sqrt{\dfrac{n-m+1}{m-1}}$$ then all the signs are reversed, so the final result is the same.

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.

• Now you need to show that these are the only solutions. – marty cohen Apr 4 '20 at 4:42
• I just want to show the two solutions of the O.P. (I was aware of what you say, even if you don't believe me). Regards. – Piquito Apr 4 '20 at 13:20