<s>[INST] Is there a cleaner way of solving this number theory problem? The problem is

If $a$ has order $3\pmod{p}$ where $p$ is an odd prime, show that $(a+1)$ has order $6\pmod{p}$.

My solution:
$$a^3-1\equiv 0\pmod{p}$$ 
$$(a-1)(a^2+a+1)\equiv 0 \pmod{p}$$
$(a-1)$ cannot be divisible by $p$ because the order of $a$ is $3$, so,
$$a^2+a+1\equiv 0\pmod{p}$$
$$(1)\qquad -a(a+1)\equiv 1\pmod{p}$$
Raising both sides to $6$, we get
$$a^6(a+1)^6\equiv 1\pmod{p}\qquad (a+1)^6\equiv 1\pmod{p}$$
To show that $6$ is the smallest integer $s$ such that $$(2)\quad (a+1)^s\equiv 1\pmod{p}$$ we use contradiciton.
From $(1)$, $(a+1)^3\equiv -1\pmod{p}$ so $s=3$ doesn't satisfy $(2)$. If $s\in\{1,2,4\}$ and $s$ satisfies $(2)$, then that would imply
$a^2\equiv 1\pmod{p}$ but that is a contradiction. If $s=5$, then
$$(a+1)^6\equiv a+1\equiv 1\pmod{p}$$
but this is impossible since $a$ is a least residue.
Is my solution correct? If so, is there a better way of solving this problem ?
 [/INST] I'll take all congruences modulo $p$.
As you say, $a^2\equiv -a-1$. Therefore $(a+1)^2\equiv -a-1+2a+1\equiv a$
and $(a+1)^3\equiv a^2+a\equiv-1$. From this, we have $(a+1)^6\equiv1$,
$(a+1)^2\not\equiv1$ and $(a+1)^3\not\equiv1$. Thus $a+1$ must have order $6$.
</s>