IMO 1967 · Problem 5

Consider the sequence \(\{c_n\}\), where \[ \begin{aligned} c_1 &= a_1+a_2+\cdots+a_8\\ c_2 &= a_1^2+a_2^2+\cdots+a_8^2\\ &\quad\cdots\\ c_n &= a_1^n+a_2^n+\cdots+a_8^n\\ &\quad\cdots \end{aligned} \] in which \(a_1,a_2,\cdots,a_8\) are real numbers not all equal to zero. Suppose that an infinite number of terms of the sequence \(\{c_n\}\) are equal to zero. Find all natural numbers \(n\) for which \(c_n=0\).

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

Answer

Precisely all odd positive integers: \[ \boxed{n=1,3,5,\ldots.} \]

Solution

For every positive integer \(m\), \[ c_{2m}=\sum_{i=1}^{8}a_i^{2m}>0, \] because every summand is nonnegative and at least one is positive. Consequently, there is a strictly increasing sequence of odd positive integers \(n_k\) such that \(c_{n_k}=0\).

Let \[ 0<r_1<r_2<\cdots<r_s \] be the distinct nonzero absolute values among \(a_1,\ldots,a_8\). For each \(j\), let \[ d_j=\#\{i:a_i=r_j\}-\#\{i:a_i=-r_j\}. \] For every odd positive integer \(n\), grouping terms with equal absolute value gives \[ c_n=\sum_{j=1}^{s}d_jr_j^n. \] Any zero values among the \(a_i\) contribute nothing.

Suppose some \(d_j\) is nonzero, and choose the largest such index \(j\). Since \(c_{n_k}=0\), division by \(r_j^{n_k}\) yields \[ 0=d_j+\sum_{i=1}^{j-1}d_i\left(\frac{r_i}{r_j}\right)^{n_k}. \] For every \(i<j\), we have \(0<r_i/r_j<1\). Thus, as \(k\to\infty\), each term in the finite sum tends to zero, giving \(d_j=0\), a contradiction. This also covers \(j=1\), when the sum is empty.

Therefore every \(d_j=0\), so \(c_n=0\) for every odd positive integer \(n\). Together with the strict positivity of all even-indexed terms, this proves the answer.

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