Let $n$ be a positive integer. Solve the system of equations \begin{align*}x_{1}+2x_{2}+\cdots+nx_{n}&= \frac{n(n+1)}{2}\\ x_{1}+x_{2}^{2}+\cdots+x_{n}^{n}&= n\end{align*} for $n$-tuples $(x_{1},x_{2},\ldots,x_{n})$ of nonnegative real numbers.
Problem
Source: Red MOP 2006
Tags: inequalities, algebra, polynomial, function, system of equations
SM4RT
17.01.2007 03:28
wow i was playing with some numbers and i noticed that if $x_{1}=x_{2}=x_{3}=...=x_{n}=1$ then: The first equation would be $1+2+3...+n=\frac{n(n+1)}{2}$ (sum of the integers from 1 to n) The second one would be $1+1+1+1...$ n times$=n$ so I dunno if thats the onyl solution but I just noticed that...
The QuattoMaster 6000
17.01.2007 03:34
Well, that has to be the only solution. Look at the second equation, there are n numbers and they all sum to n. The only way to make n positive integers sum to n is to have them all equal to one.
chess64
17.01.2007 03:37
$x_{i}$ are not necessarily integers.
t0rajir0u
17.01.2007 03:46
Let $a_{i}= ix_{i}$. Then $\sum a_{i}$ is fixed by the first equation, so we can use Karamata on the second to deduce that $a_{i}= i$ is the only solution.
I think.
Farenhajt
17.01.2007 03:55
It's easy to notice that $x_{1}=\dots=x_{n}=1$ is a solution. Hence substitute $x_{1}=1+y_{i}$ where $y_{i}\geqslant-1$. Then the equations become
$y_{1}+2y_{2}+\dots+ny_{n}=0$
$(1+y_{1})+(1+y_{2})^{2}+\dots+(1+y_{n})^{n}=n$
Assume at least one of $y_{i}$ is $\neq-1,0$. Then by Bernoulli's inequality we have that the LHS of the last equation is strictly greater than $1+y_{1}+1+2y_{2}+\dots+1+ny_{n}$, which yields $y_{1}+2y_{2}+\dots+ny_{n}<0$, and that contradicts the first equation.
(Note: If $y_{1}\neq-1,0$ and all the other $y_{i}$ are $=-1,0$, Bernoulli can't be applied directly, but then it's easy to see that the system has no solution.)
Therefore, $(\forall i)y_{i}\in\{-1,0\}\implies(\forall i)x_{i}\in\{0,1\}$. Now it's obvious that $x_{1}+2x_{2}+\dots+nx_{n}$ can reach the desired value of ${n(n+1)\over 2}$ iff $x_{i}=1$. Hence that's the only solution.
SplashD
17.01.2007 03:57
chess64 wrote: Let $n$ be a positive integer. Solve the system of equations \begin{align*}x_{1}+2x_{2}+\cdots+nx_{n}&= \frac{n(n+1)}{2}\\ x_{1}+x_{2}^{2}+\cdots+x_{n}^{n}&= n\end{align*} for $n$-tuples $(x_{1},x_{2},\ldots,x_{n})$ of nonnegative real numbers. The second equation minus the first equation equals $(x_{2}-1)^{2}+(x_{3}-1)^{2}(x_{3}+2)+\cdots+(x_{n}-1)^{2}(x^{n-2}+2x^{n-3}+3x^{n-4}+\cdots+(n-1))=0$ and the polynomials that have all postive integer coefficients clearly can't have postive real roots, and thus the only solution is $n$ $1s$
t0rajir0u
17.01.2007 04:16
Haha, those are both much cooler than what I did.
SplashD
17.01.2007 04:26
What function would you have used for your karamata solution?
chess64
17.01.2007 04:33
Are you talking about the Karamata/Majorization Inequality?
t0rajir0u
17.01.2007 04:40
Huh. Yeah, I was, but I don't think it works anymore. Intuitively, what I was thinking of amounts to Farenhajt's solution, so...
parmenides51
11.04.2020 14:48
another link and a source (Czech-Polish-Slovak 2005), as I am creating the 2006 MOP Homework post collection