2009/10 AHSMC Part 1 Problem 15

If $p = 2$, $2^p + p^2 = 8$ is not prime. If $p = 3$, $2^p + p^2 = 17$ is prime. If $p > 3$, then $p = 6k \pm 1$ for some integer $k$, so that $p^2 = 36k^2 \pm 12k + 1$ is $1$ more than a multiple of $3$. On the other hand, when divided by $3$, successive powers of $2$ leave remainders of $2$ and $1$ alternately. Since $p > 3$ is odd, $2^p$ is $2$ more than a multiple of $3$. Hence $2^p + p^2$ is divisible by $3$, and cannot be prime. The answer is (B).