A question was asked recently on this site:
Find bounds on $\ \displaystyle\sum_{i=1}^{n} {a_i} ^5\ $ if $\ \displaystyle\sum_{i=1}^{n} a_i = 0,\ \sum_{i=1}^{n} {a_i}^3 = 0\ $ and $\ \displaystyle\sum_{i=1}^{n} \lvert a_i \rvert = 1.$
I then asked for one example of $a_i$ that satisfy the three conditions in the question, and the user correctly gave the example: $a_1 = -0.5,\ a_2 = 0.5 $ and then he deleted the question. Now, the only other examples I can think of which satisfy all three conditions are $\left( \underbrace{\frac{-1}{2n},\ \frac{-1}{2n},\ \ldots,\ \frac{-1}{2n}}_{n\ \text{times}},\ \underbrace{\frac{1}{2n},\ \frac{1}{2n},\ \ldots,\ \frac{1}{2n}}_{n\ \text{times}} \right).\ $ This works for any $\ n\in\mathbb{N}.$
My question is:
Are there any other examples satisfying all three conditions?
I found some examples that convince me there are numbers that satisfy $\displaystyle\sum_{i=1}^{n} {a_i}^3 = 0\ $ and $\ \displaystyle\sum_{i=1}^{n} \lvert a_i \rvert = 1,\ $ for example $\ 0.40,\ -0.375, -0.225\ $ is close enough to convince me. However, these don't come close to satisfying the condition $\ \displaystyle\sum_{i=1}^{n} a_i = 0.$
Similarly, there are many numbers satisfying $\ \displaystyle\sum_{i=1}^{n} a_i = 0\ $ and $\ \displaystyle\sum_{i=1}^{n} \lvert a_i \rvert = 1:\ $ The positive terms must sum to $0.5$ and the negative terms must sum to $-0.5,$ but other than this there are no restrictions.
And as an example where $\ \displaystyle\sum_{i=1}^{n} {a_i}^3 = 0\ $ and $\ \displaystyle\sum_{i=1}^{n} a_i = 0,\ $ there's $\left( \underbrace{\frac{-1}{2n},\ \frac{-1}{2n},\ \ldots,\ \frac{-1}{2n}}_{n\ \text{times}},\ \underbrace{\frac{1}{2n},\ \frac{1}{2n},\ \ldots,\ \frac{1}{2n}}_{n\ \text{times}} \right),$ but I'm sure there's other examples also. For example, $(-0.2,-0.3, 0.5)$ doesn't work because ${0.5}^3 > \vert {-0.2}^3 + {-0.3}^3\vert,\ $ however we can split the $0.5$ up into three (or more) positive numbers $a_i, i=3,\ldots, n,\ $ such that $ \displaystyle\sum_{i=3}^{n} a_i = 0.5\ $ where $n\geq 5.$ Since $\ \displaystyle\sum_{i=3}^{n} {a_i}^3\ $ has min value $\ n \left( \frac{0.5}{n} \right)^3\ $ (I think), and due to continuity of the multi-variable function $\ \displaystyle\sum_{i=3}^{n} {a_i}^3,\ $ there are clearly examples such that $ \vert {-0.2}^3 + {-0.3}^3\vert = \displaystyle\sum_{i=3}^{n} {a_i}^3.$
We can prove that there are infinitely many “non-trivial” examples.
First choose an integer $m \ge 4$ and positive distinct real numbers $b_1, b_2, \ldots, b_m$. For sufficiently small $d \in \Bbb R\setminus \{ 0 \}$ the polynomial equation $$ (x-b_1)(x-b_2)\cdots (x-b_m) - d = 0 $$ has $m$ distinct positive real solutions $c_1, c_2, \ldots, c_m$, which are all different from the $b_j$. Then $$ (x-b_1)(x-b_2)\cdots (x-b_m) - d = (x-c_1)(x-c_2) \cdots (x-c_m)\, . $$ By Vieta's formulas is $e_K(b1, \ldots, b_m) = e_K(c_1, \ldots, c_m)$ for the first three elementary symmetric polynomials $e_1, e_2, e_3$: $$ \sum_{j=1}^m b_j = \sum_{j=1}^m c_j \, , \, \sum_{1 \le j < k \le m}^m b_jb_k = \sum_{1 \le j < k \le m}^m c_jc_k \, , \, \sum_{1 \le j < k < l\le m}^m b_jb_kb_k = \sum_{1 \le j < k < l\le m}^m c_jc_kc_l \, . $$
By Expressing power sums in terms of elementary symmetric polynomials it follows that $p_k(b1, \ldots, b_m) = p_k(c_1, \ldots, c_m)$ for the first three power sums $p_1, p_2, p_3$. In particular is $$ \sum_{j=1}^m b_j = \sum_{j=1}^m c_j \, , \quad \sum_{j=1}^m b_j^3 = \sum_{j=1}^m c_j^3 \, . $$ Then $$ (a_1, \ldots, a_{2n}) = (-b_1, \ldots, -b_m, c_1 \ldots, c_m) $$ satisfies $\sum_{j=1}^{2n} a_j = 0$ and $\sum_{j=1}^{2n} a_j^3 = 0$. The condition $\sum_{j=1}^{2n} |a_j| = 1$ can then be obtained by scaling.
If we choose $b_1 = 0$ in the above construction then with a suitable choice of the sign of $d$, the $c_j$ are still positive, that gives examples with odd $n = 2m-1$.
Example: $x(x-1)(x-2)(x-3)+ 0.2 =0$ has the solutions $$ (c_1, c_2, c_3, c_4) \approx (0.0356137, 0.9037007, 2.0962993,2.9643863) $$ so that $(a_1, \ldots, a_7) = (-1, -2, -3, c_1, c_2, c_3, c_4)$ satisfy $\sum_{j=1}^{7} a_j = 0$ and $\sum_{j=1}^{7} a_j^3 = 0$.