<s>[INST] If $a$ has order 3 $\pmod p$, $p$ prime, then $1+a+a^2 \equiv 0 \pmod p$ and $1+a$ has order $6$ There is a little proposition left  to the readers in my number theory book, and I'm not sure how to answer this in a proper way:

If $a$ has order $3 \pmod p$, $p$ prime, then $1+a+a^2 \equiv 0 \pmod p$ and $1+a$ has order $6$.

Some hints?
 [/INST] How $a$ has order $3$, then $a\not\equiv 1$.
Now, note that $(a-1)(1+a+a^2) = a^3-1 \equiv 0$ (mod $p$). As $\mathbb{Z}_p$ is a field and $a-1\neq 0$, so $1+a+a^2=0$ (mod $p$).
To see why $1+a$ has order $6$, note that 
$$(1+a)^6=(1+2a+a^2)^3=a^3=1$$
So, the order of $1+a$ divide $6$.
If $1+a$ has order $1$, then $1+a=1\Rightarrow a=0$. This contradicts the fact that $a$ has order $3$.
If $1+a$ has order $2$, then $(1+a)^2=1\Rightarrow 1+2a+a^2=1\Rightarrow a=1$. But $1$ has order $1$, not $3$.
If $1+a$ has order $3$, then $(1+a)^3=1\Rightarrow (1+a)(1+2a+a^2)=1\Rightarrow (1+a)a=1\Rightarrow a+a^2=1\Rightarrow 1+a+a^2=2\Rightarrow 0=2$, so $p=2$. But $\mathbb{Z}_2$ has no elements with order $3$.
So, the only possible order for $1+a$ is $6$.
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