 # Understand the problem

For $n\ge3$, determine all real solutions of the system of $n$ equations :

$x_1+x_2+\cdots+x_{n-1}=\frac{1}{x_n}$

………………………….

$x_1+x_2+\cdots+x_{i-1}+x_{i+1}+\cdots+x_{n-1}+x_n=\frac{1}{x_i}$

…………………………..

$x_2+x_3+\cdots+x_{n-1}+x_n=\frac{1}{x_1}$.

##### Source of the problem

I.S.I. (Indian Statistical Institute) B.Math.(Hons.) Entrance Examination 2008. Subjective Problem no. 9.

##### Topic

System of $n$ Equations

9 out of 10

##### Suggested Book

Mathematical Olympiad Challenges by Titu Andreescu & Razvan Gelca.

Do you really need a hint? Try it first!

Let $s = x_1+x_2+\cdots+x_i+\cdots +x_{n-1}+x_n$.

Then the system of $n$ equations are equivalent to $x_i^2-sx_i+1=0$ for $i=1,2,….,n$.

It follows that $x_1,x_2,…..,x_{n-1},x_n$ are solutions to the quadratic equation: $u^2-su+1=0$.                                                   $\cdots\cdots\cdots\cdots (i)$

Now we have two possible cases.

Case – I :  Two roots of the equation (i) are equal, then , all $x_i$ are equal.

i.e. , $x_1=x_2=\cdots=x_{n-1}=x_n=u$.

$\Rightarrow s=nu$.

Putting this value of $s$ in equation (i) we get, $(n-1)u^2=1$.

$\Rightarrow u=\frac{1}{\pm \sqrt{n-1}}$.

Case – II : Two roots of equation (i) are not equal. Then let these two roots are $u_1$ and $u_2$. Also let among {$x_1,x_2,……,x_{n-1},x_n$}   $k$  of them equal to $u_1$ and $(n-k)$ of them equal to $u_2$, where $0<k<n$.

In this case, we have , (a) $u_1+u_2=s$.    [Sum of roots of equation (i)]

(b) $u_1\cdot u_2=1$.    [ Product of roots of equation (i)].

Now,.   $s=x_1+x_2+\cdots+x_{n-1}+x_n=k\cdot u_1+(n-k)\cdot u_2$

$\Rightarrow s=(u_1+u_2)+(k-1)\cdot u_1+(n-k-1)\cdot u_2$

$\Rightarrow (k-1)\cdot u_1+(n-k-1)\cdot u_2=0$   [using (a)]

$\Rightarrow (k-1)\cdot u_1^2=(k+1-n)\cdot u_1\cdot u_2=k+1-n\le 0$ [using (b) and since ,$k<n\Rightarrow k+1\le n$].

$\Rightarrow u_1^2\le 0$

But $u_1\neq 0$ as the coefficient of $u^0$ is 1($\neq 0$) in equation (i).

$\Rightarrow u_1^2<0$

$\Rightarrow u_1$ is not real. So we have no real solution to the system of $n$ equations in this case.

Thus the total no. of real solutions to the system of $n$ equations is two and these two solutions are:

$x_1=x_2=\cdots=x_{n-1}=x_n=\frac{1}{\sqrt{n-1}}$      and

$x_1=x_2=\cdots=x_{n-1}=x_n=-\frac{1}{\sqrt{n-1}}$.(Ans.)

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