Let $p \equiv 1(n)$. Let $\chi$ denote a character from $\mathbb{F}_p \rightarrow \mathbb{C}-\{0\}$. Then, in Ireland and Rosen's A Classical Introduction to Number Theory, while trying to find the number of solutions to $X^n+Y^n=1$, the authors claim that if $-1$ is a $n^{th}$ power, the sum $\sum_{j=0}^{n-1}\chi^j(-1)$ is $n$, and $0$ otherwise. I'm not sure why this follows?
My work so far:
In the case that $-1$ isn't an $n^{th}$ power, I am guessing that the sum is zero as $\chi^j(-1)$ must be a $p-1^{th}$ root of unity and then as $n \mid p-1$, the sum of all roots of unity is $0$.
If $-1$ is an $n^{th}$ power, then there exists some $a$ such that $a^n \equiv -1 \mod p$ (is this what it means?). Then,
\begin{align*} \sum_{j=0}^{n-1}\chi^{j}(-1)&=\sum_{j=0}^{n-1}\chi^j(a^n)=\sum_{j=0}^{n-1}\chi^j(a)^n=\sum_{j=0}^{n-1}\chi^{jn}(a) \ldots \end{align*}
and then I do not how to proceed. Any help would be appreciated, thank you so much!
Edit: In response to comments, I have attached the relevant screenshots below:
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