I would like to solve the following system of algebraic equations:
$$ \sum_{i=1}^N {\alpha_i}=0 \\ \sum_{i=1}^N{\alpha_i\beta_i}=0\\ \sum_{i=1}^N{\alpha_i\beta_i^2}=0 $$ where $\alpha_i,\beta_i \in \mathbb{Z}$ and $N>3$ and $\alpha_i\neq 0\; \forall i$.
Remark:
For $N=2$: no non trivial solutions,
For $N=3$, if $\alpha=(\alpha_i)_i$ is non zeros then it is orthogonal to $(1,1,1)$, to $\beta=(\beta_i)_i$ and to $\gamma=(\beta_i^2)_i$.
Therefore $\det((1,1,1),\beta,\gamma)=\prod_{1\le i<j\le N}(\beta_j-\beta_i)=0$
Then for instance $\beta_1=\beta_2=b$ which leads to :
$b=0$ or $\beta_3=b$ or ($\alpha_1=-\alpha_2$ and $\alpha_3=0$) ... and the problem seems intricate
- For $N>3$, no idea
For $N\ge 4$ there exists an additional solution different from the "obvious ones". For this it is enough to consider $N=4$. We obtain the rational solution \begin{align} \alpha_1 & =\frac{\alpha_4( - \beta_2\beta_3 + \beta_2\beta_4 + \beta_3\beta_4 - \beta_4^2)}{\beta_1^2 - \beta_1\beta_2 - \beta_1\beta_3 + \beta_2\beta_3}, \\[0.2cm] \alpha_2 & =\frac{\alpha_4(\beta_1\beta_3 - \beta_1\beta_4 - \beta_3\beta_4 + \beta_4^2)}{\beta_1\beta_2 - \beta_1\beta_3 - \beta_2^2 + \beta_2\beta_3}, \\[0.2cm] \alpha_3 & = \frac{\alpha_4( - \beta_1\beta_2 + \beta_1\beta_4 + \beta_2\beta_4 - \beta_4^2)}{\beta_1\beta_2 - \beta_1\beta_3 - \beta_2\beta_3 + \beta_3^2}, \end{align} with arbitrary $\beta_i$ such that the denominators are nonzero.
A typical "obvious solution" is given by $$ \alpha_2= - \alpha_3,\; \beta_1=\beta_4, \; \beta_2=\beta_3,\; \alpha_4=-\alpha_1. $$