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

Difficulty Level

9 out of 10

Suggested Book

Mathematical Olympiad Challenges by Titu Andreescu & Razvan Gelca.

Start with hints

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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