# 1979 IMO Problems/Problem 5

## Problem

Determine all real numbers a for which there exists non-negative reals $x_{1}, \ldots, x_{5}$ which satisfy the relations $\sum_{k=1}^{5} kx_{k}=a,$ $\sum_{k=1}^{5} k^{3}x_{k}=a^{2},$ $\sum_{k=1}^{5} k^{5}x_{k}=a^{3}.$

## Solution

Discussion thread can be found here: [1]

Let $\Sigma_1= \sum_{k=1}^{5} kx_{k}$, $\Sigma_2=\sum_{k=1}^{5} k^{3}x_{k}$ and $\Sigma_3=\sum_{k=1}^{5} k^{5}x_{k}$. For all pairs $i,j\in \mathbb{Z}$, let $$\Sigma(i,j)=i^2j^2\Sigma_1-(i^2+j^2)\Sigma_2+\Sigma_3$$ Then we have on one hand $$\Sigma(i,j)=i^2j^2\Sigma_1-(i^2+j^2)\Sigma_2+\Sigma_3=\sum_{k=1}^5(i^2j^2k-(i^2+j^2)k^3+k^5)x_k =\sum_{k=1}^5k(i^2j^2-(i^2+j^2)k^2+k^4)x_k$$ Therefore \\(1) $$\Sigma(i,j)=\sum_{k=1}^5k(k^2-i^2)(k^2-j^2)x_k$$ and on the other hand \\ (2) $$\Sigma(i,j)=i^2j^2a-(i^2+j^2)a^2+a^3=a(a-i^2)(a-j^2)$$ Then from (1) we have$$\Sigma(0,5)=\sum_{k=1}^5k^3(k^2-5^2)x_k\leq 0$$ and from (2) $$\Sigma(0,5)=a^2(a-25)$$ so $a\in [0,25]$ Besides we also have from (1) $$\Sigma(0,1)=\sum_{k=1}^5k^3(k^2-1)x_k\geq 0$$ and from (2) $$\Sigma(0,1)=a^2(a-1)\geq 0 \implies a\notin (0,1)$$ and for $n=1,2,3,4$ $$\Sigma(n,n+1)=\sum_{k=1}^5k(k^2-n^2)(k^2-(n+1)^2)x_k$$ where in the right hand we have that $$k, so $\[(k^2-n^2)(k^2-(n+1)^2)>0$$, $$k=n,n+1 , \implies (k^2-n^2)(k^2-(n+1)^2)=0$$ and $$k>n \implies (k^2-n^2)(k^2-(n+1)^2)>0$$, so $$\Sigma(n,n+1)\geq 0$$ for $n=1,2,3,4$ From the latter and (2) we also have $$\Sigma(n,n+1)=a(a-n^2)(a-(n+1)^2))\geq 0\implies a\notin (n^2,(n+1)^2)$$ So we have that $$a\in [0,25]-\bigcup_{n=0}^4(n^2,(n+1^2))=\{0,1,4,9,16,25\}$$
If $a=k^2$, $k=0,1,2,3,4,5$ take $x_k=k$, $x_j=0$ for $j\neq k$. Then $\Sigma_1=k^2=a$, $\Sigma_2=k^3k=k^4=a^2$, and $\Sigma_3=k^5k=k^6=a^3$