If $g$ is a primitive $p$-root of unity, then $g^n\equiv 1 \pmod p \iff (p-1)|n$

Question

This statement is from Caratheodory's Theory of Functions of a complex variable vol.1 p.282:

Let $s_n(p):=0^n+1^n+2^n+...(p-1)^n$. Then if $p$ is a prime and $g$ is any number that doesn't divide $p$, we have:

$$g^ns_n(p) \equiv s_n(p) \pmod p$$

The two statements that I am after are:

If $g$ is a primitive $p$-root of unity, then $g^n\equiv 1 \pmod p \iff (p-1)|n$. Hence if $n$ is not divisible by $(p-1)$ we have $s_n(p)\equiv 0 \pmod p$.

How am I supposed to reduce mod $p$ the power of a a primitive $p$-root of unity? In $\Bbb F_p$ the only $p$-root of unity is $1$...

Related : http://math.stackexchange.com/questions/560861/sums-of-primitive-roots-and-quadratic-residues-when-p-equiv-3-pmod-4 — lab bhattacharjee, Jan 18 '14 at 13:40
http://math.stackexchange.com/questions/433678/sums-of-powers-below-a-prime — lab bhattacharjee, Jan 18 '14 at 13:49

score 1 · Accepted Answer · answered Jan 18 '14 at 13:45

Taking the Question in the title,

Let Multiplicative order ord$_pa=d \implies d$ is the smallest positive integer such that $a^d\equiv1\pmod p$

Now the $\displaystyle a^m\equiv1\pmod p$ and $m=d\cdot e+r$ where $0\le r<d$

So, $\displaystyle a^m=a^{d\cdot e+r}=(a^d)^e\cdot a^r$

$\displaystyle \implies 1\equiv1^e\cdot a^r\pmod p\implies a^r\equiv1$

So, we find another non-negative integer $r<d$ such that $\displaystyle a^r\equiv1\pmod p$

$\implies r=0\implies d|m$

Now as $g$ is a primitive root, ord$_pg=\phi(p)=p-1$

If $g$ is a primitive $p$-root of unity, then $g^n\equiv 1 \pmod p \iff (p-1)|n$

1 Answers1