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 $2$.

For an odd prime $q \geq 5$ we have $q^2 \equiv 3$. And if $q = 2n + 1$ then $2^q = (2^n)^2 \dot 2 \equiv 2$. So $2^q + q^2 \equiv 1$.
