Divisibility rules with prime number Let $n \in \mathbb{Z}$ with the property that
$$
7 \mid\left(n^{3}+1\right)
$$
but $7$ does not divide $\left(n^{2}-2 n-3\right) .$ Prove that $7 \mid(4 n+1)$.
So far I got:
$n^3+1 = (n+1)\cdot(n^2-n+1)$, since $7$ is a prime, so
$7\mid(n+1)$ or $7\mid(n^2-n+1)$.
And since $7$ does not divide $\left(n^{2}-2 n-3\right) = (n-3)\cdot(n+1)$, I know that $7\mid(n^2-n+1)$.
And $7$ does not divide the difference of $(n^2-n+1)$ and $(n^2-2n-3)$ which is $n+4$.
Same since $7$ is prime, $7$ does not divide $4n+16\Rightarrow$$7$ does not divide $4n+2$.
Then I am stuck here, how could I use the information I got so far to prove $7 \mid(4 n+1)$?
Really hope someone could help/hint me with it! Thank you!

Two methods:
Method 1
If $n^3\equiv-1\bmod 7$ then $n\in\{3,5,6\}\bmod 7$. But $n\in\{3,6\}$ renders $n^2-2n-3=(n+1)(n-3)\equiv0$, denied by hypothesis, and the last possibility $n\equiv5$ renders $4n+1\equiv0$.
Method 2
Multiply $n^2-2n-3$ by $4n+1$:
$(n^2-2n-3)(4n+1)=4n^3-7n^2-14n-3\equiv4(n^3+1)\bmod 7$
Thus if the prime number $7$ divides $n^3+1$ it must divide (at least) one of the factors $n^2-2n-3$ or $4n+1$.
