# $g^{0p^nk}+g^{p^nk}+g^{2p^nk}+…+g^{(p-2)p^nk} \equiv 0$ (mod $p^n)$ if $p-1$ doesn't divide $k$

Let $$p$$ be prime and $$n\geq 1$$. Let $$g$$ be the integer equivalent of some generator for $$(\mathbb{Z}/p\mathbb{Z})^ \times$$.

Let $$k\in \{0,1,2,...\} \subseteq \mathbb{Z}$$ such that $$p-1$$ does NOT divide $$k$$.

I want to show that $$g^{0p^nk}+g^{p^nk}+g^{2p^nk}+...+g^{(p-2)p^nk} \equiv 0$$ (mod $$p^n)$$

I tried writing the sum as $$g^{0p^nk}+g^{p^nk}+g^{2p^nk}+...+g^{(p-2)p^nk}=\frac {g^{(p-1)p^nk}-1}{g^{p^nk}-1}$$ but I don't see why this would have to be divisible by $$p^n$$

• What's $n$? Taking $p=3=k$ we get $0^3+1^3+2^3=9$ so, yes, it is divisible by $3^2$, but not by $3^3$. – lulu Nov 9 '18 at 0:18
• @lulu You're right, I must have made a mistake. I hope I fixed the question now. – Pascal's Wager Nov 9 '18 at 0:36
• Looks like it is about the multiplicative order of $g^{p^n k}$ in $(\mathbb{Z}/p^n\mathbb{Z})^ \times$ being a non-trivial divisor of $p-1$ – random Nov 9 '18 at 14:08

1. The multiplicative group $$(\Bbb Z/p)^\times$$ ist cyclic of order $$p-1$$. So $$g^{(p-1)k} \equiv 1$$ mod $$p$$.
2. On the other hand, the condition that $$k$$ does not divide $$p-1$$ implies that $$g^{p^n k} \equiv g^k \not \equiv 1$$ mod $$p$$, in other words, the denominator of your fraction is not divisible by $$p$$.
3. As you just asked in $$a \equiv b$$ (mod $$p$$) implies $$a^{p^n} \equiv b^{p^n}$$ (mod $$p^n$$)?, the congruence from 1. further implies that $$g^{(p-1)k p^n} \equiv 1^{p^n} = 1$$ mod $$p^n$$. In other words, the numerator of your fraction is ...