Skip to main content
Source Link
Amir
  • 303
  • 11

Only for $m=2$, $\sum_{i=1}^mx^{2n+1}_k$ is always divisible by $\sum_{i=1}^mx_k$. Indeed,

$$x+y | x^{2n+1}+y^{2n+1}, n \in \mathbb N.$$

Post Made Community Wiki by Amir