Proof using exhaustion $n^4 - 1$ is divisible by $5$ where $n$ is not divisible by $5$. The title pretty much states it.
Proof using exhaustion $n^4 - 1$ is divisible by $5$ where $n$ is not divisible by $5$. 
Can anyone give me a hint how to approach this?
Do I need to consider separate case where n is odd and when n is even?
Solution: use division algorithm

We have $n^4-1 = (n-1)(n+1)(n^2+1)$. Hence, if
\begin{align}
n \equiv 5k+1 & \implies 5 \vert n-1  & \implies 5 \vert n^4-1\\
n \equiv 5k-1 & \implies 5 \vert n+1 & \implies 5 \vert n^4-1\\
n \equiv 5k\pm2 & \implies 5 \vert n^2+1 & \implies 5 \vert n^4-1
\end{align}
