Proof that $3\mid n^3 − 4n$

Question

Prove that $n^3 − 4n$ is divisible by $3$ for every positive integer $n$.

I am not sure how to start this problem. Any help would be appreciated

Answer 1

Well, one can start off by factoring $n^3 - 4n$:

$n^3 - 4n = n(n^2 - 4)$ $= n(n + 2)(n - 2); \tag{1}$

next we note that every $n \in \Bbb Z$ is of one of the three forms

$n = 3q, \tag{2}$

$n = 3q + 1, \tag{3}$

or

$n = 3q + 2; \tag{4}$

this follows from the basic properties of Euclidean division applied to $n$ with $3$ as the divisor; the only possible remainders $r$ be $0$, $1$, and $2$, since such integers $r$ must satisfy $0 \le r \le 2$. In case (2), clearly $3 \mid n$; in case (3), we have

$n + 2 = 3q + 1 + 2$ $= 3q + 3 = 3(q + 1), \tag{5}$

showing that $3 \mid n + 2$; finally, case (4) yields

$n - 2$ $= 3q + 2 - 2 = 3q, \tag{6}$

whence $3 \mid n - 2$. For evey possible $n$, $3$ divides one of the factors of $n^3 - 4n$ occurring in (1); thus we have $3 \mid n^3 - 4n$ for all $n \in \Bbb Z$. QED.

Answer 2

$n^3-n-3n = n(n^2-1)-3n = n(n-1)(n+1)-3n$. As, $3|3n$ and $3|(n-1)$ or $3|n$ or $3|(n+1) \implies 3|n^3-4n$

Answer 3

Note that $$ \begin{align} n^3-4n &=6\binom{n}{3}+6\binom{n}{2}-3\binom{n}{1}\\ &=3\left[2\binom{n}{3}+2\binom{n}{2}-\binom{n}{1}\right] \end{align} $$