The sum of odd powered real numbers equals zero implies the numbers are inverses 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 $a_i+a_j=0$.
 A: 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.
