How to solve this quadratic congruence equation? Well, we have :
$$n^2+n+2+5^{4n+1}\equiv0\pmod{13}$$
i'm little bit confused, I think i can solve this using the reminders of $n^2$, $n$ and $5^{4n+1}$ over $13$, by the way I have no idea about the Chinese Reminder Theorem no need to use it. and thanks in advance
edit:
$4 \le n \le 25$

It is easy to see that 
$$5^2\equiv -1 \pmod{13}$$
So, we have 
$$5^4\equiv 1 \pmod{13}$$
Therefore, for any n, we have 
$$5^{4n+1}\equiv 5\pmod{13}$$
So, the equation simplifies to
$$n^2+n+7\equiv 0\pmod{13}$$
Considering vieta, we check the factors of 7,20,33,...
 Looking at the factors of 20 , we notice that 2 and 10 sum upto 12, which
 is $\equiv -1 \pmod{13}$, so 2 and 10 are the desired solutions.
