# Modulo arithmetic and sum of arbitrary powers of a primitive root of unity.

Prove that if $w$ is a primitive nth root of unity, then

$1 + w^k + (w^k)^2 + (w^k)^3 + \cdots + (w^k)^{n-1} =0$ iff $k \neq 0$ mod $n$.

Sorry for the terrible formatting. Also, I don't know anything about groups and rings, so please keep the answer as elementary as possible! Thanks.

Note : $k$ is a non-zero integer.

• This is the sum of a finite geometric progression. Nov 9, 2017 at 22:05

## 1 Answer

Hint: $(1-x)(1+x+x^2+\ldots+x^{n-1}) = 1 - x^n$.

• Hmm.. Then 1+x+x^2.. +x^n-1 = 1-x^n /1-x We replace x by w^k So 1+w^k .... = 1-w^nk / 1-x 1+w^k.... =0 iff 1-w^nk = 0, which is only possible if w^nk = 1. But w^nk = 1 if and only if k is a multiple of n, i.e k = 0 mod n because w is a primitive root.
– John
Nov 9, 2017 at 22:04
• Indeed, as long as $x \ne 1$. Nov 9, 2017 at 22:05
• Isn't that contradicting k =/=0 mod n?
– John
Nov 9, 2017 at 22:09
• Nevermind, I see what I did wrong. Thanks!
– John
Nov 9, 2017 at 22:27