Show that the odd prime divisors of the integer $n^2+n+1$ which are different from $3$ are of the form $6k+1$. 1)The odd prime divisors of the integer $n^4+1$ are of the form $8k+1$.
My attempt: Let $p$ be odd divisor of $n^4+1$.Then $n^4+1 \equiv 0$ (mod p) $\Rightarrow n^4 \equiv -1$ (mod p) $\Rightarrow n^8 \equiv 1$ (mod p). Hence, the order of $n$ modulo $p$ is 8, which implies that $8 | \phi(p)=p-1 \Rightarrow p=8k+1$
The part i not sure is the order of $n$ modulo $p$. How do I show the order of $n$ modulo $p$ is 8?
2)Show that the odd prime divisors of the integer $n^2+n+1$ which are different from $3$ are of the form $6k+1$.
My attempt: Let $p$ be odd divisor of $n^2+n+1$. Then $n^2+n+1 \equiv 0 $(mod p). I try to mimick the method above but i fail. 
Can anyone guide me?

Hint: $$4(n^2+n+1)=(2n+1)^2+3$$
Alternatively:
$$(n-1)(n^2+n+1)=n^3-1$$
