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 $2$.

To show that congruence of $2^q\pmod 3$ is equal to 2 for a prime $q\geq 5$, it's sufficient to prove it for odd numbers, because any prime $q\geq 5$ is odd. So it reduces to verifying that $2^1\equiv 2\pmod 3$  and $2^2\equiv 1\pmod 3$.
