If $n > 1$ is an integer not of the form $6k + 3$, prove that $n^{2} + 2^{n}$ is composite. I try to divide in to $5$ case $6k, 6k+1, 6k+2, 6k+4, 6k+5$
and put them in $n^{2}$ and consider the possible last digit 0f $2^n$
and sum with those form
I can only prove $6k, 6k+2, 6k+4$ case .the last digit is even number and can divide by $2$.
But I can't prove $6k+1, 6k+5$ case because might be odd and not guarantee Ican divide by some numbers 
Anyone can help me by give a hint ?

The hint is $n > 1$. When $n = 1$, $n^2 + 2^n = 3$ which is prime. This hints that $n^2 + 2^n$ may be divisible by $3$.
We know that $6k+1$ can be written as $3(2k)+1$ and $6k+5$ can be written as $3(2k+1)+2$. Thus if $n$ is of the form $6k+1$ or $6k+5$, then $n^2 \equiv 1\pmod3$. 
The next part is to prove that $2^n \equiv 2 \pmod 3$ if $n$ is of such form. Notice that $2^0 \equiv 1 \pmod 3$, $2^1 \equiv 2 \pmod 3$ and $2^2 \equiv 1 \pmod 3$. Thus for all $k \ge 0$, $2^{2k+1} \equiv 2 \pmod 3$. Since $6k+1$ can be written as $2(3k)+1$ and $6k+5$ can be written as $2(3k+2)+1$, $2^n \equiv 2 \pmod 3$.
Thus  $n^2 + 2^n \equiv 1+2 \pmod 3 \equiv 0 \pmod 3$. This implies that $n^2 + 2^n$ is divisible by 3 if $n = 6k+1$ or $n = 6k+5$.
