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How would I check when the expression $$\Sigma_{i=1}^{j-1} (1+\zeta_j^i)^n$$ is equal to $0$, where all $\zeta_j^i$ are the $j$th roots of unity not equal to $1$?

Of course, then I would be looking for an expression of $n$ in terms of $j$ or vice versa.

I have a suspicion that looking at this geometrically may help but I'm not sure how. I also bashed out the cases for $j=3$ and $j=4$ and found that $j=3$ has no solutions and $j=4$ has solutions of $n=3,7\mod 8$.

qt.
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  • do you mean j=m ? – zwim Mar 25 '17 at 04:09
  • Whoops, fixed it – qt. Mar 25 '17 at 04:13
  • http://math.stackexchange.com/questions/72129/how-did-they-simplify-this-expression-involving-roots-of-unity, this helps a little but does not solve the problem, since we still have a sum in the end. And the characterization in term of binomial sum does not help either, I think. – zwim Mar 25 '17 at 04:56

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