Primitive root modulo $p$.

Question

Let $a$ be an integer and $p$ an odd prime number. Suppose $\bar{a}$ has order $h>1$ in $\mathbb{Z}_p^*$, how can we show that $$a^{h-1} + a^{h-2} + ... + a + 1\equiv 0\mod p?$$

Answer 1

Hint : You have that $$(a-1)(a^{h-1}+a^{h-2}+\dots +1)=a^h-1.$$ Hence $(a-1)(a^{h-1}+a^{h-2}+\dots +1)\mod p=0$.

Answer 2

Use the high school factorisation $$x^n-1=(x-1)(x^{n-1}+x^{n-2}+\dots+x+1).$$ Note the relation is valid only if $a\not\equiv 1\mod p$.