Is it true that :
$((2^n+n^2) \in \mathbf{P} \land n \geq 3)\Rightarrow n\equiv 0 \pmod 3 $
I have checked this statement for the following consecutive values of $n$ : $3,9,15,21,33,2007,2127,3759$
Note that $2^n+n^2$ is special case of the form $2^n+k\cdot n$. One can easily show using Dirichlet's theorem that for any odd $n$ sequence $a_k=2^n+k\cdot n$ contains infinitely many primes.
Any idea how to prove or disprove statement above ?