1979 IMO Problems/Problem 5

Problem

Determine all real numbers a for which there exists positive 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]

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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 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 $k<n \implies (k^2-n^2)<0, (k^2-(n+1)^2)<0$ , 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 later 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 $\Sigma_3=k^5k=k^6=a^3$

1979 IMO (Problems) • Resources
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 6
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1979 IMO Problems/Problem 5

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