Prove that $2^q+q^2$ is divisible by 3 where $q$ is a prime and $q\geq5$. I'm looking to prove that $2^q+q^2$ is divisible by $3$ where $q$ is a prime such that $q\geq5$.
I know that primes greater than five will be congruent to either $1\ (\text{mod}\ 3)$ or $2\ (\text{mod}\ 3)$, which means that the $q^2$-term will always be congruent to $1\ (\text{mod}\ 3)$ which simplifies the problem to finding the congruence of $2^q+1\ (\text{mod}\ 3)$. However, I'm unable to go any further as I'm unable to show that the congruence of $2^q\ (\text{mod}\ 3)$ is always equal to $2$.

$2^{1} \equiv 2 \pmod{3}$;
$2^{2} \equiv 1 \pmod{3}$;
$2^{3} \equiv 2 \pmod{3}$;
$2^{4} \equiv 1 \pmod{3}$;
$2^{5} \equiv 2 \pmod{3}$;
$2^{6} \equiv 1 \pmod{3}$;
$...$
Note that when the exponent is odd, the remainder equals to $2$ and when the exponent is even, the remainder is $1$. 
This pattern keeps repeating because multiplying number with a remainder of $1\pmod{3}$ by $2$ equals to $2\pmod{3}$, and multiplying again by $2$ gives us the remainder of $4$, which is $1\pmod{3}$.
Because $q$ is odd, $2^{q} \equiv 2 \pmod{3}$.
