Prove, for any positive integer $n$, that $n -3$ must be a multiple of $5$ if $n^3 -n -4$ is a multiple of $5$. I had previously solved the problem of proving that $n^3-n-4$ must be a multiple of $5$, given that $n-3$ is a multiple of $5$.  I did so by algebraically manipulating $n^3-n-4$ into:
$$
2(n-3)+(n-1)(n-1)((n-3)+5)
$$
Given that the first term is a given multiple of $5$, and the second term is a product of a multiple of $5$, I could prove directly that the sum of these terms was a multiple of $5$.
With the new (reverse) problem, I can't do this direct algebraic manipulation, and I believe the way to go about this problem is proof by cases, where the list of exhaustive cases would be when the remainder is $0$, $1$, $2$, $3$, or $4$.  
My question is, what steps should I take to set up and show the proof for cases?  I am new to these numerical theory problems, and so any basic guidance would be very much appreciated.

Write $n$ in the form $5k+r$, where $r\in\{0,1,2,3,4\}$. Show that $n^3-r^3$ is a multiple of $5$, as of course is $n-r$, so that
$$(n^3-n-4)-(r^3-r-4)=(n^3-r^3)-(n-r)$$
is a multiple of $5$. Thus, if $n^3-n-4$ is a multiple of $5$, so is $r^3-r-4$. Now show that $3$ is the only value of $r$ for which $r^3-r-4$ is a multiple of $5$.
