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Show that if real numbers $a_1,a_2,\ldots,a_n$ satisfy $$a_1^l+a_2^l+\cdots+a_n^l=0$$ for every odd $l$, then for any $a_i$ we can always find some $a_j$ (not necessarily different) such that $a_i+a_j=0$.

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    $\begingroup$ What's your problem? $\endgroup$ – Shuchang Sep 10 '13 at 9:08
  • $\begingroup$ My guess: “any odd $l$” should be “every odd $l$”, and the conclusion should be $a_i+a_j=0$. And I would solve it by starting with the $a_i$ with the largest absolute value and considering the limit $l\to\infty$. $\endgroup$ – Harald Hanche-Olsen Sep 10 '13 at 9:12
  • $\begingroup$ Harald you guess is right. I corrected my question. Can you elaborate on how to go about the problem by starting with the $a_i$ with the largest absolute value? Are we expecting to get a contradiction? $\endgroup$ – Kuai Sep 10 '13 at 10:34
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In principle we do an induction on $n$. The result is obviously true at $n=1$ and $n=2$.

Without loss of generality we may assume that none of the $a_i$ is $0$. For if some $a_i$ is equal to $0$, we can remove it, reducing the problem to the case $n-1$.

Without loss of generality we may assume that if $a_i$ and $a_j$ have the same absolute value, they are in fact equal. For if $a$ and $-a$ both occur in the sequence, they can be paired and removed, and we are at the case $n-2$.

It will make things easier if we use the fact that the problem is scale-invariant: The result holds for the numbers $a_1,a_2,\dots,a_n$ if and only if it holds for the numbers $Ca_1,Ca_2,\dots,Ca_n$, whatever non-zero constant $C$ we choose.

Arrange the numbers in non-decreasing order of absolute value. Call them $b_1,b_2,\dots,b_n$.

By scale invariance, we may assume that $b_n=1$. Several other $b_i$ may be equal to $1$, say $k$ of them.

Let the largest absolute value smaller than $1$ that occurs among the absolute values of the $b_i$ be $\delta\lt 1$. Then $$b_n^l +b_{n-1}^l +b_{n-2}^l +\cdots +b_1^l \ge k(1^l) -(n-k)\delta^l.\tag{1}$$ We show that for large enough $l$, we have $$k-(n-k)\delta^l \gt 0.\tag{2}$$ That, together with Inequality (1), will contradicts the assumption that $a_1^l+a_2^l +\cdots+a_n^l=0$ for all odd $l$.

To prove Inequality (2), we need to show equivalently that $$\delta^l \lt \frac{k}{n-k}$$ for large enough $l$. This is clear, since $\frac{k}{n-k}$ is positive, and $\lim_{l\to\infty} \delta^l=0$.

Remark: Basically, it comes down to the fact that for large enough $l$, $a_1^l+a_2^l+\cdots+a_n^l$ is dominated by the terms of largest absolute value. The above inequalities pin down the details.

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  • $\begingroup$ Why the down vote? $\endgroup$ – nbubis Sep 10 '13 at 17:25
  • $\begingroup$ Presumably the downvoter did not like it. There is reason not to, it is overkill. $\endgroup$ – André Nicolas Sep 10 '13 at 17:28
  • $\begingroup$ This is clear, thanks! $\endgroup$ – Kuai Sep 10 '13 at 20:39
  • $\begingroup$ You are welcome. $\endgroup$ – André Nicolas Sep 10 '13 at 20:43
  • $\begingroup$ @AndréNicolas Can it be proven for complex numbers?? $\endgroup$ – Astor Aug 17 '17 at 22:41

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