Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.8 AIME Find the answer

57. (USS 4) ${ }^{\mathrm{IMO} 5}$ Consider the sequence $\left(c_{n}\right)$ : $\begin{gathered} c_{1}=a_{1}+a_{2}+\cdots+a_{8}, \\ c_{2}=a_{1}^{2}+a_{2}^{2}+\cdots+a_{8}^{2}, \\ \cdots \\ \cdots \cdots \cdots \\ c_{n}=a_{1}^{n}+a_{2}^{n}+\cdots+a_{8}^{n}, \end{gathered}$ [^1]where $a_{1}, a_{2}, \ldots, a_{8}$ are real numbers, not all equal to zero. Given that among the numbers of the sequence $\left(c_{n}\right)$ there are infinitely many equal to zero, determine all the values of $n$ for which $c_{n}=0$ .

Your answer

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Solution

57. Obviously $c_{n}>0$ for all even $n$ . Thus $c_{n}=0$ is possible only for an odd $n$ . Let us assume $a_{1} \leq a_{2} \leq \cdots \leq a_{8}$ : in particular, $a_{1} \leq 0 \leq a_{8}$ . If $\left|a_{1}\right|n_{0}, 7\left|a_{1}\right|^{n}7 a_{1}^{n}+a_{8}^{n}>0$ , contradicting the condition that $c_{n}=0$ for infinitely many $n$ . Similarly $\left|a_{1}\right|>\left|a_{8}\right|$ is impossible, and we conclude that $a_{1}=-a_{8}$ . Continuing in the same manner we can show that $a_{2}=-a_{7}, a_{3}=-a_{6}$ and $a_{4}=-a_{5}$ . Hence $c_{n}=0$ for every odd $n$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.