Prove that $\operatorname{ord}_{3^{2n}+3^n+1}2 \equiv 0 \pmod{4}$ 
Prove that $\operatorname{ord}_{3^{2n}+3^n+1}2 \equiv 0 \pmod{4}$ where $n$ is a positive integer.

For $n = 1$ we have $3^{2n}+3^n+1 = 3^2+3+1 = 13$. In this case, since $12$ is prime we know that the order must divide $12$. We find that $12$ is the order. 
For $n = 2$ we have $3^{2n}+3^n+1 = 3^4+3^2+1 = 91$. In this case, $91$ is not prime but we find that the order is $12$.
In both of these cases, the order was $12$ and is divisible by $4$.
For the general case, I thought about first proving that $\operatorname{ord}_{3^{2n}+3^n+1}2$ is even. We have $$3^{2n}+3^n+1 = (3^n+1)^2-3^n.$$ How can we prove it is even?

Clearly, it's sufficient to find divisor $d$ of $3^{2n}+3^n+1$ such that $\operatorname{ord}_d 2 \equiv 0 \pmod 4$.
If $n$ is even, say $n=2k$, then 
$$
3^{2n} + 3^n + 1 = (3^{k})^4 + (3^k)^2 + 1 = 
\bigl([3^k]^2+3^k+1\bigr)\bigl([3^k]^2-3^k+1\bigr),
$$
it is divisible by $3^{2k}+3^k+1$. 
And when $n$ is odd, then
$$
3^{2n} + 3^n + 1 = (3^n - 1)^2 + 3^{n+1}
$$
is sum of two coprime squares, hence all its prime divisors are congruent to $1$ modulo 4 (that's obvious if you're familiar with gaussian integers). Moreover,
$$
3^{2n} + 3^n + 1 \equiv 1 + 3 + 1 = 5 \pmod 8,
$$
it means that it has prime divisor $p \equiv 5 \pmod 8$.
And $2$ is quadratic non-residue modulo that $p$, it means that
$$
2^{(p-1)/2} \equiv -1 \pmod p.
$$
Put for clarity $p = 4k+1$. Now it's easy to see that $\operatorname{ord}_p 2  \equiv 0 \pmod 4$: $\operatorname{ord}_p 2  $ divides $(p-1) = 4k$ but not divides $(p-1)/2 = 2k$.
