A number theory problem by muhammad manaf

$\begin{aligned} n^3 + 6 & \equiv 0 \text{ (mod n+2)} \\ ({\color{#3D99F6}(n+2)}-2)^3 + 6 & \equiv 0 \text{ (mod n+2)} \\ {\color{#3D99F6}(n+2)}^3 - 3{\color{#3D99F6}(n+2)}^2(2) + 3{\color{#3D99F6}(n+2)}(2^2) - 2^3 + 6 & \equiv 0 \text{ (mod n+2)} \\ - 8 + 6 & \equiv 0 \text{ (mod n+2)} \\ - 2 & \equiv 0 \text{ (mod n+2)} \end{aligned}$

Since the factors of $-2$ are $-2,-1,1,2$ , we have $n+2 = \begin{cases} - 2 & \implies n = -4 \\ -1 & \implies n=-3 \\ 1 & \implies n=-1 \\ 2 & \implies n=0 \end{cases}$ , a total of $\boxed{4}$ $n$ 's that satisfy the congruence.

$n^3+6\equiv0\pmod{n+2}$ $n^3+6\equiv n^3+2n^2\pmod{n+2}$ $2n^2\equiv6\pmod{n+2}$ $2n^2\equiv2n^2+4n+6\pmod{n+2}$ $4n+6\equiv0\pmod{n+2}$ $4n+6\equiv4n+8\pmod{n+2}$ $2\equiv0\pmod{n+2}$

So,n+2 is the factor of 2 (the negative too)

$n+2=\{-2,-1,1,2\}$ $n=\{-4,-3,-1,0\}$ there is $\boxed{4}$ n's satisfying the equation