# Prove that if the $\mathrm{ord}(p)a =3$, then $\mathrm{ord}(p)(a+1) = 6$, $p$ prime [duplicate]

I couldn't answer this question. This only one. It looks simple, but I got stuck. Here is the image of the question. The number $12$. Sorry for my english.

Prove that if the $\mathrm{ord}(p)a=3$, then $\mathrm{ord}(p)(a+1)=6$

Example: $3^3=1(\mathrm{mod}\;13)$ then $4^6=1(\mathrm{mod\;}13)$.

Prove that if the $\mathrm{ord}(p)a=3$, then $\mathrm{ord}(p)(a+1)=6$

## marked as duplicate by Dietrich Burde, Juniven, Daniel W. Farlow, kingW3, user26857Apr 4 '17 at 16:24

Hint: We have $a^3 - 1=(a-1)(a^2+a+1)$.