# Let $w \in G_{15}$ be a primitive root. Find every $n \in \mathbb{N}$ such that $\sum_{i=2}^{n-1} w^{3i} = 0$

We can first rewrite the series in a useful form,

$$\sum_{i=2}^{n-1} w^{3i} = \bigg( \sum_{i=0}^{n-1} w^{3i} \bigg) - w^3 - 1$$

But since $$w$$ is primitive, we can apply the geometric series formula,

$$\frac{w^{3n} - 1}{w^3 - 1} - w^3 - 1 = 0 \ \iff \ w^{n} = w^2$$

So, the answer should be $$n=2$$?

Let $$w$$ be a primitive $$15$$-th root of unity.
For $$n\le 2$$, the relation $$\sum_{k=2}^{n-1} w^{3k} = 0$$ holds trivially.
Suppose $$n\in\mathbb{N}$$, with $$n > 2$$.
Then as you derived, $$\sum_{k=2}^{n-1} w^{3k} = 0\;\iff\;\frac{w^{3n}-1}{w^3-1}-w^3-1=0$$
Continuing from that result, \begin{align*} &\frac{w^{3n}-1}{w^3-1}-w^3-1=0\\[4pt] \iff\;&\frac{w^{3n}-1}{w^3-1}=w^3+1\\[4pt] \iff\;&w^{3n}-1=(w^3+1)(w^3-1)\\[4pt] \iff\;&w^{3n}-1=w^6-1\\[4pt] \iff\;&w^{3n}=w^6\\[4pt] \iff\;&w^{3n-6}=1\\[4pt] \iff\;&3n-6\equiv 0\;(\text{mod}\;15)\\[4pt] \iff\;&3n\equiv 6\;(\text{mod}\;15)\\[4pt] \iff\;&n\equiv 2\;(\text{mod}\;5)\\[4pt] \end{align*} hence, the set of qualifying values of $$n$$ is $$\{1\}\cup \{n\in\mathbb{N}\mid n\equiv 2\;(\text{mod}\;5)\}$$