Why is $n^2+4$ never divisible by $3$? Can somebody please explain why $n^2+4$ is never divisible by $3$?  I know there is an example with $n^2+1$, however a $4$ can be broken down to $3+1$, and factor out a three, which would be divisible by $4$.
Write $n=3r+q$ with $q\in\{0,1,2\}$. Then
$$
n^2+4=9r^2+6rq+q^2+4=(9r^2+6rq+3)+q^2+1.
$$
The expression in the parentheses is divisible by $3$ whereas inspecting the three possibilities for $q$ reveals that $q^2+1$ is never divisible by $3$.
Hint: What does $-4$ equal to in $\mathbb{Z_3}$?