Find the remainder of $2^n+n^2$ modulus 6 
Find the remainder of $2^n+n^2$ modulus 6 given that $2^n+n^2$ is a prime and $n\geq2$($n$ positive integer)  

I tried to solve this but failed!I just know that $n$ must be odd. No progress at all!!

$n$ can't be even because then $n^2+2^n$ is even and larger than $2$.
Therefore $n$ is odd and $2^n\equiv 2\bmod 3$. Notice that $n^2$ can only be $1$ or $0\bmod 3$.  In the first case $3$ divides $2^n+n^2$ and $n>3$, implying $2^n+n^2$ is not prime.
We conclude $n^2\equiv 0\bmod 3$. Therefore $2^n+n^2$ is odd and $2\bmod 3$. So $2^n+n^2\equiv 5\bmod 6$
