How to solve this quadratic congruence equation?

Question

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:

$4 \le n \le 25$

Answer 1

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.

Answer 2

I'm very sorry about this bad resolution i've used a webcam, i'm going to update it later on.