Prove n^2+2^n Composite if n not 6k+3

[fresh_42]

You can write the even case without a $c$. If $n$ is even, then $n=2k\, , \,k>0$ and $n^2+2^n=4k^2+4^k\equiv 0 \mod 2$ which is a bit shorter.

There are often situations in mathematics where symmetric cases (like odd/even) are not treated by the same methods. One case could be easy, the other one difficult, or require more sub-cases like $c=1,5$ in our case.