Is $\sum_{j=1}^{k}a^j \equiv 0 \mod p\;$? where k is the order of $a \mod p$, with $p$ being an odd prime?

Question

In other words is $a^1 + a^2 + \dotsm a^k \equiv 0\mod p\;$?

This is true when $a$ is a primitive root of $p$ because $a^1, a^2, \dotsc a^k$ are congruent to $1,2,\dotsc,p-1$ in some order. Hence, the sum is a triangular number of form $\frac{(p-1)(p)}{2}$ which is clearly divisible by p, i.e. is $0 \mod p$.

However the question is does this work for all $a$ $\textbf{not}$ being primitive root of $p$? E.g. take $p=11, a=3$.

We have $3^1\equiv3$, $3^2\equiv9$, $3^3\equiv5$, $3^4\equiv4$, $3^5\equiv1$, with the sum of these $\equiv22\equiv0 \mod11$.

If this is true, how would one go about proving this?

Much appreciated.

Answer 1

What if $k=1\iff a=1?$

Else using Geometric Series $$\sum_{r=1}^ka^r=\frac{a(a^k-1)}{a-1}$$

Multiplicative Order $\displaystyle k\implies a^k\equiv1\pmod p\iff a^k-1\equiv0\implies a(a^k-1)\equiv0$

$\displaystyle\implies \frac{a(a^k-1)}{a-1}\equiv0\pmod{\frac p{\text{gcd}(a-1,p)}}$ (using $\#12$ of this)

If $\displaystyle(a-1,p)=1$ for which, "$p$ is a prime" is a sufficient condition

$\displaystyle\implies \frac{a(a^k-1)}{a-1}\equiv0\pmod p$

Answer 2

As pointed out in the comments, the claim is false for $a=1$ (because $1\ne0$), and nonsensical for $a=0$ (becasue $0$ hasn't got multiplicative order). So, let's assume $a\ne0,1$ (here is where the assumption $p$ odd prime needs) and call $S$ your sum. Now (recall that $a^k=1$): \begin{alignat}{1} aS &= a(1+a+\dots+a^{k-1}) \\ &= a+a^2+\dots+a^k \\ &= 1+a+a^2+\dots \\ &= S \\ \tag1 \end{alignat} Since $\mathbb Z/p\mathbb Z$ is a field, from $(1)$ follows that, if $S\ne0$, then $a=1$: contradiction. Hence $S=0$. (So, actually, there's no need for the geometric partial sum formula.)