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# 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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