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 $a_i+a_j=0$.

  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.