Show that if $a$ has order $3\bmod p$ then $a+1$ has order $6\bmod p$. Show that if $a$ has order $3\bmod p$ then $a+1$  has order $6\bmod p$. I know I am supposed to use primitive roots but I think this is where I am getting caught up. The definition of primitive root is "if $a$ is a least residue and the order of $a\bmod p$ is $\phi(m)$ then $a$ is a primitive root of $m$. But I really am not sure how to use this to my advantage in solving anything.
Thanks!
 A: Note that if $a=1$, $a$ has order $1$. Thus, we can assume $a\ne1$. Furthermore, $p\ne2$ since no element mod $2$ has order $3$. Therefore, $-1\ne1\pmod{p}$.
$$
\begin{align}
(a+1)^3
&=a^3+3a^2+3a+1\\
&=1+3a^2+3a+1\\
&=-1+3(1+a+a^2)\\
&=-1+3\frac{a^3-1}{a-1}\\
&=-1
\end{align}
$$
Therefore, $(a+1)^3=-1$ and $(a+1)^6=1$. (Shortened a la ccorn). 
$$
\begin{align}
(a+1)^2
&=a^2+2a+1\\
&=a+(a^2+a+1)\\
&=a+\frac{a^3-1}{a-1}\\
&=a\\
&\ne1
\end{align}
$$
Therefore, $(a+1)^2\ne1$.
Thus, $(a+1)$ has order $6$.
A: ord_$pa=3\iff p$ divides $(a^3-1)=(a-1)(a^2+a+1)$
But $p\not\mid (a-1)$ as ord_$pa=3 $
$\displaystyle\implies a^2+a+1\equiv0\pmod p$
$\displaystyle\iff a(a+1)\equiv-1\pmod p$
Method $1:$
$\implies a^3(a+1)^3\equiv-1\pmod p\implies (a+1)^3\equiv-1$ and $(a+1)^6\equiv1$
$\implies $ord_$p(a+1)\mid 6$ but does not divide $3$
If ord_$p(a+1)\mid 2, (a+1)^2\equiv1\pmod p$ and $a^2(a+1)^2\equiv1\pmod p\implies a^2\equiv1$ which is impossible as ord$_pa=3$
$\implies $ord$_p(a+1)=6$
Method $2:$
$\displaystyle a(a+1)\equiv-1\pmod p\iff a+1\equiv (-a)^{-1}\pmod p$
Again as $a^3\equiv1\pmod p, (-a)^3=-a^3\equiv-1\pmod p\implies (-a)^6\equiv1\implies $ord$_p(-a)=6$
Using this,  ord$_p\{(-a)^{-1}\})=$ord$_p(-a)=6$
