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

Question

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.

$\begingroup$ This is the sum of a finite geometric progression. $\endgroup$
– Angina Seng
Nov 9, 2017 at 22:05 — Angina Seng, Nov 9, 2017 at 22:05

xyzzyz · Accepted Answer · 2017-11-09 22:02:27Z

1

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

answered Nov 9, 2017 at 22:02

xyzzyz

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$\begingroup$ 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. $\endgroup$
– John
Nov 9, 2017 at 22:04
$\begingroup$ Indeed, as long as $x \ne 1$. $\endgroup$
– xyzzyz
Nov 9, 2017 at 22:05
$\begingroup$ Isn't that contradicting k =/=0 mod n? $\endgroup$
– John
Nov 9, 2017 at 22:09
$\begingroup$ Nevermind, I see what I did wrong. Thanks! $\endgroup$
– John
Nov 9, 2017 at 22:27

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