Mediterranean Mathematical Competition

Since the data of the problem concern $n$ real numbers $x_1,x_2,\dots ,x_n$ in the closed interval $[-4,2]$, we consider the polynomial $$P(x)=(x+4)(x-2)(x-1{)}^2,$$ which in $[-4,2]$ satisfies the relation $$P(x)=(x+4)(x-2)(x-1{)}^2\le 0.(1)$$ Adding by parts the inequalities coming from (1) for $x=x_1,x_2,\dots ,x_n$, and taking in mind the conditions of the problem, we find: $$0\ge P(x_1)+\cdots +P(x_n)=\sum _{i=1}^n{x}_i^4-11\sum _{i=1}^n{x}_i^2+18\sum _{i=1}^n{x}_i-8n\ge 34n-11\cdot 4n+18n-8n=0.(2)$$ Hence, since $P(x_i)\le 0$, for all $i=1,2,\dots ,n$, from relation (2) we have: $P(x_i)=0$, for all $i=1,2,\dots ,n$, which means that $x_i\in \{-4,1,2\}$, for all $i=1,2,\dots ,n$. We suppose that from the integers $x_1,x_2,\dots ,x_n$, $a$ are equal to $-4$, $b$ are equal to $1$ and $c$ are equal to $2$. Then we have $a+b+c=n$ and from the data of the problems we have the inequalities $$\left\{\begin{array}{l}-4a+b+2c\ge a+b+c\\ 16a+b+4c\le 4(a+b+c)\\ 256a+b+16c\ge 34(a+b+c)\end{array}\right\}\iff \left\{\begin{array}{l}c\ge 5a\\ b\ge 4a\\ 222a\ge 33b+18c\end{array}\right\}.$$ By multiplying both parts of the first inequality with $18$ and the second with $33$ and summing the produced inequalities by parts we get the inequality $$33b+18c\ge 132a+90a=222a,$$ which in combination with the inequality $222a\ge 33b+18c$ gives: $$33b+18c=222a,$$ which is valid, if and only if $b=4a$, $c=5a$, that is $$a+b+c=10a\text{ or }10a=n.$$ Therefore the numbers $x_1,x_2,\dots ,x_n$ there exist, if and only if, $n$ is a multiple of $10$. For $n=10m$, where $m$ is a positive integer a possible solution arises by taking $m$ times the number $-4$, $4m$ times the number $1$ and $5m$ times the number $2$.