Q:

Let $o_p(a) = 3$, show that $o_p(1+a) = 6$

The question is : let $o_p(a) = 3$. Show that $o_p(1+a) = 6.$
I'm not sure of how to solve this question and would appreciate any help. Thank you in advance.

A:

I will assume $p$ is prime.
Since $ord_p(a)=3$ we have $p\mid a^3 -1 = (a-1)(a^2+a+1)$. Since
$ord_p(a) >1$ we have $p\nmid a-1$ and thus $p\mid a^2+a+1$. Now $$(a+1)^6-1 \equiv_p21(a^2+a+1)\equiv_p 0$$ so $ord_p(1+a) \mid 6$. Now show $ord_p(1+a) >3$...

