IMO 1979 · Problem 5

Find all real numbers \(a\) for which there exist non-negative real numbers \(x_1,x_2,x_3,x_4,x_5\) satisfying the relations \[ \sum_{k=1}^{5}kx_k=a,\qquad \sum_{k=1}^{5}k^3x_k=a^2,\qquad \sum_{k=1}^{5}k^5x_k=a^3. \]

Official text of the problem. Official problem paper (PDF)

Answer

\[ \boxed{a\in\{0,1,4,9,16,25\}.} \]

Solution

The first relation implies \(a\geq 0\), since all \(x_k\) are non-negative. If \(a=0\), all three relations hold with \(x_1=\cdots=x_5=0\).

Now suppose \(a>0\). Using the three given relations, we obtain \[ \begin{aligned} \sum_{k=1}^{5}kx_k(k^2-a)^2 &=\sum_{k=1}^{5}k^5x_k -2a\sum_{k=1}^{5}k^3x_k +a^2\sum_{k=1}^{5}kx_k\\ &=a^3-2a\cdot a^2+a^2\cdot a\\ &=0. \end{aligned} \] Every summand is non-negative, so each must be zero. Moreover, since \(\sum_{k=1}^{5}kx_k=a>0\), some \(x_m>0\). For this index, \[ mx_m(m^2-a)^2=0 \] forces \(a=m^2\). Thus \(a\in\{1,4,9,16,25\}\).

Conversely, for any \(m\in\{1,2,3,4,5\}\), set \(a=m^2\), \(x_m=m\), and \(x_k=0\) for \(k\ne m\). Then \[ \sum_{k=1}^{5}kx_k=m^2=a,\qquad \sum_{k=1}^{5}k^3x_k=m^4=a^2,\qquad \sum_{k=1}^{5}k^5x_k=m^6=a^3. \] Hence every listed value is attainable, completing the proof.

The solution was checked against John Scholes' IMO solution archive (Kalva).