Prove that $2^p+p^2$ is prime for $p=3$ only I do know that all prime numbers larger than $3$ can be expressed as $3k + 1$ and or $3k + 2$. Plugging those in I still see no solution.
EDIT: $p$ can only be a prime number.

Whether $p\equiv 1\pmod 3$ or $p\equiv 2\pmod 3$, either way $p^2\equiv 1\pmod 3$
Since $p$ must be odd, $2^p\equiv 2\pmod 3$
Adding them together gets you a number that is divisible by $3$.
