IMO 1967 Problem 5

Proposed by: -
Verified: no
Verdicts: PASS + FAIL
Solve time: 15m08s

Problem

Consider the sequence ${ c_n }$, where $$ c_1 = a_1 + a_2 + \cdots + a_8 $$ $$ c_2 = a_1^2 + a_2^2 + \cdots + a_8^2 $$ $$ \cdots $$ $$ c_n = a_1^n + a_2^n + \cdots + a_8^n $$ $$ \cdots $$ 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 $c_n = 0$.

Exploration

Testing small configurations clarifies what structure is forced by infinitely many vanishing power sums. If $a$ and $-a$ appear with equal multiplicity, then $a^n+(-a)^n$ vanishes exactly for odd $n$ and contributes $2a^n$ for even $n$, which is always positive when $n$ is even. Adding further symmetric pairs of different magnitudes, such as ${1,-1,2,-2}$, shows that even powers produce strictly positive sums because every term becomes $|a_i|^n$, while odd powers cancel pairwise.

Trying an asymmetric multiset, such as ${1,1,-1}$, already produces only finitely many zeros, since dominant terms eventually prevent cancellation. Any attempt to mix unmatched magnitudes fails similarly: once a largest magnitude is unbalanced, its contribution cannot be removed by smaller terms infinitely often. These checks suggest that cancellation infinitely often forces symmetry at each magnitude level.

The same behaviour persists under further tests with $n=1,2,3,4,5$, where even exponents eliminate signs and produce strictly positive contributions unless all terms are arranged in cancelling pairs.

This points toward a layered symmetry condition by magnitude, rather than a single global reduction.

Problem Understanding

Real numbers $a_1,\dots,a_8$ are not all zero, and

$c_n = \sum_{i=1}^8 a_i^n.$

It is given that $c_n=0$ for infinitely many natural numbers $n$. The task is to determine all natural numbers $n$ for which $c_n=0$ must hold.

The condition constrains the structure of the multiset ${a_i}$ so strongly that it forces a rigid pattern in the exponents at which complete cancellation can occur.

Key Observations

Let $M$ be the maximum of $|a_i|$ among the nonzero terms. The contributions from elements of magnitude $M$ dominate all others after normalization by $M^n$, so any infinite cancellation pattern must already be visible in the top magnitude layer.

For indices with $|a_i|=M$, each term can be written as $a_i = M\varepsilon_i$ where $\varepsilon_i \in {-1,1}$. The contribution of this layer is $M^n\sum \varepsilon_i^n$. Since $\varepsilon_i^n=\varepsilon_i$ when $n$ is odd and $\varepsilon_i^n=1$ when $n$ is even, the parity structure is rigid.

Infinite vanishing forces the top layer contribution to vanish infinitely often, which can only happen if the number of $+M$ and $-M$ terms is equal. Removing these paired terms reduces the problem to a smaller maximum magnitude, and repeating the argument enforces the same symmetry at every magnitude level.

Thus every nonzero value appears in a pair $(x,-x)$ with equal multiplicity.

Once this pairing structure is established, the behavior of $c_n$ reduces to sums of expressions of the form $x^n+(-x)^n$.

Solution

Let $a_1,\dots,a_8$ be real numbers, not all zero, and assume that $c_n=\sum_{i=1}^8 a_i^n$ vanishes for infinitely many natural numbers $n$.

Let $M$ be the maximum of $|a_i|$ among the nonzero $a_i$. Consider the set $I={i:|a_i|=M}$. Write each $a_i$ with $i\in I$ as $a_i=M\varepsilon_i$ where $\varepsilon_i\in{-1,1}$. Then

$\sum_{i\in I} a_i^n = M^n \sum_{i\in I} \varepsilon_i^n.$

If $n$ is even, this equals $M^n|I|$. If $n$ is odd, this equals $M^n(#{\varepsilon_i=1}-#{\varepsilon_i=-1})$.

Let $(n_k)$ be an infinite sequence such that $c_{n_k}=0$. Dividing by $M^{n_k}$ gives that the contribution of indices outside $I$ tends to $0$, so

$\sum_{i\in I} \varepsilon_i^{n_k} \to 0.$

If infinitely many $n_k$ are even, then $\sum_{i\in I} \varepsilon_i^{n_k}=|I|$, forcing $|I|=0$, impossible. Hence infinitely many $n_k$ are odd, and for those indices the expression becomes a constant integer independent of $n_k$. Therefore that constant must be zero, giving equal numbers of $+M$ and $-M$ terms.

Remove all terms of magnitude $M$. The remaining sequence still has infinitely many zero power sums, since subtracting the vanishing contribution of the top layer preserves the property. Repeating the argument finitely many times yields that every nonzero $a_i$ occurs in a pair $(x,-x)$.

Now write the multiset of nonzero values as ${\pm x_1,\dots,\pm x_m}$ with equal multiplicity in each pair. Then for every $n$,

$c_n = \sum_{j=1}^m \bigl(x_j^n + (-x_j)^n\bigr).$

If $n$ is even, each term satisfies $x_j^n + (-x_j)^n = 2x_j^n>0$, so $c_n>0$ and cannot vanish.

If $n$ is odd, each term satisfies $x_j^n + (-x_j)^n=0$, so $c_n=0$.

Thus $c_n=0$ holds for all and only odd natural numbers $n$.

$\boxed{\text{All odd natural numbers } n}$

Verification of Key Steps

The reduction to the maximal magnitude layer is justified because after scaling by $M^n$, all smaller magnitude contributions converge to zero, so along any subsequence where the total sum is exactly zero, the dominant layer must itself approach zero. Since the dominant layer takes only finitely many integer-valued possibilities depending on parity, convergence forces eventual exact cancellation structure in that layer.

The elimination of unmatched $+M$ and $-M$ terms follows from checking both parities: even indices produce a strictly positive contribution from the top layer, contradicting infinitely many zeros, so infinitely many vanishing indices must be odd, forcing equality of counts in the top layer.

Inductive removal of magnitude levels is valid because after subtracting symmetric pairs at the current maximal magnitude, the remaining set still produces infinitely many zeros, allowing repetition of the same argument finitely many times.

For the final structure, even exponents convert each symmetric pair into a strictly positive contribution, ruling out cancellation, while odd exponents annihilate each pair exactly, producing vanishing sums for all odd $n$.

Alternative Approaches

A structural alternative uses the fact that $(c_n)$ is a linear recurrence sequence with characteristic roots $a_i$, so its zero set is a finite union of arithmetic progressions. Since the sequence is real and involves no periodic complex phases other than sign changes, the only possible infinite zero progression is the set of odd integers, arising from symmetry of roots under $x\mapsto -x$.