Prove that $S_k(p)= \begin{cases} -1, & \text{if $(p-1)\mid k$ } \\ 0, & \text{otherwise} \end{cases}$ in $\Bbb Z_p$

Question

Consider $\Bbb Z_p= \Bbb Z/{p \Bbb Z}$ where $p$ is an odd prime.

Now denote $S_k(p)= \sum _{j=0}^{p-1} j^k$ in $\Bbb Z_p$

The problem is to prove that $S_k(p)= \begin{cases} -1, & \text{if $(p-1)\mid k$ } \\ 0, & \text{otherwise} \end{cases}$ in $\Bbb Z_p$

Now we know $\Bbb Z_p^*$ is a cyclic group of order $p-1$, so in the first case if we have $k=a(p-1)$, then $\forall j \in \Bbb Z_p^*$ we have $j^k=j^{(p-1)a}=1$. Hence $S_k(p)=p-1=-1$ in $\Bbb Z_p$

if $(p-1)$ does not divide $k$ then I am having problem. If $\gcd(p-1,k)=1$ then $S_k(p)= \sum _{j=0}^{p-1} j=p(p-1)/2=0$ in $\Bbb Z_p$

But for the other cases?? Please help and explain in details.

Answer 1

Lemma(I): Let $p$ to be an odd prime, then $\big( \mathbb{Z}_p^* , . \big)$ is a cyclic group.

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Lemma(II): For every $i \in \mathbb{Z}_p^*$; we have: $i \mathbb{Z}_p^* = \mathbb{Z}_p^*$.

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Lemma(III): Let $p$ to be an odd prime, and let $k$ to be any arbitrary integer.

• If $p-1 \mid k$; then for every $j \in \mathbb{Z}_p^*$; we have: $j^k \overset{p}{\equiv} 1$.

• If $p-1 \nmid k$; then there exists $l \in \mathbb{Z}_p^*$; such that: $l^k \overset{p}{\ncong} 1$.

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Proof:

• The first statement is the trivial result of fermat's little theorem.

• Second statement: Let $\varepsilon$ to be a generator of $\big( \mathbb{Z}_p^* , . \big)$;
  and let $d:=\gcd(k,p-1)$.
  Also let $t:=\dfrac{p-1}{d}$;
  then it is easy to check that for every $r$, with $\gcd(r,t)=1$;
  we have: $(\varepsilon^r)^k \overset{p}{\ncong} 1$.

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Lemma(IV): Let $G$ be any group and $a \in G$ any element of finite order. Then we have: $$\text{ord}(a^t)=\dfrac{\text{ord}(a)}{\gcd(\text{ord}(a),t)}.$$

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• If $p-1 \mid k$; then for every $j \in \mathbb{Z}_p^*$; we have: $j^k \overset{p}{\equiv} 1$. So the sum $ {\sum}_{j \in \mathbb{Z}_p^*}j^k= {\sum}_{j \in \mathbb{Z}_p^*} 1= p-1=-1.$

• If $p-1 \nmid k$; then there exists $l \in \mathbb{Z}_p^*$; such that: $l^k \overset{p}{\ncong} 1$. Notice that: $$ \color{Blue}{ {\sum}_{ j \in \mathbb{Z}_p^*} j^k }= {\sum}_{lj \in \mathbb{Z}_p^*} (lj)^k = {\sum}_{ j \in \mathbb{Z}_p^*} l^kj^k = \color{Red}{l^k} \color{Blue}{ {\sum}_{ j \in \mathbb{Z}_p^*} j^k } \\ \Longrightarrow (1-\color{Red}{l^k} ) \color{Blue}{ {\sum}_{ j \in \mathbb{Z}_p^*} j^k } = 0 ; $$ but notice that $(1-\color{Red}{l^k} )$ is not zero;
  which implies that: $\color{Blue}{ {\sum}_{ j \in \mathbb{Z}_p^*} j^k }=0$.