Question

Prove that for any positive integer $n,n^3-n$ is divisible by $6$.

Answer 1

Step I: To prove the divisibility by $3$.

When a number is divided by $3$, the possible remainders are $0\mathrm{or}1\mathrm{or}2$.

$\therefore$ $n=3p\mathrm{or}3p+1\mathrm{or}3p+2$, where $p$ is some integer.

Case 1: Consider $n=3p$

Then $n$ is divisible by $3$.

Case 2: Consider $n=3p+1$

Then $n–1$ $=3p+1–1$

$\Rightarrow n-1=3p$ is divisible by $3$.

Case 3: Consider $n=3p+2$

Then $n+1=3p+2+1$

$\Rightarrow n+1=3p+3$

$\Rightarrow n+1=3(p+1)$ is divisible by $3$.

So, we can say that one of the numbers among $n,n–1\mathrm{and}n+1$ is always divisible by $3$.

$\Rightarrow n(n–1)(n+1)$ is divisible by $3$.

Step II: To prove the divisibility by $2$.

Similarly, when a number is divided by $2$, the possible remainders are $0\mathrm{or}1$.

$\therefore$$n=2q\mathrm{or}2q+1$, where $q$ is some integer.

Case 1: Consider $n=2q$

Then $n$ is divisible by $2$.

Case 2: Consider $n=2q+1$

Then $n–1=2q+1–1$

$n–1=2q$ is divisible by $2$ and

$n+1=2q+1+1$

$n+1=2q+2$

$n+1=2(q+1)$ is divisible by $2$.

So, we can say that one of the numbers among $n,n–1\mathrm{and}n+1$ is always divisible by $2$.

$\therefore$ $n(n–1)(n+1)$ is divisible by $2$.

Step III: To prove the divisibility by $6$

Since, $n(n–1)(n+1)$ is divisible by $2\mathrm{and}3$.

Therefore, as per the divisibility rule of $6$, the given number is divisible by $6$.

Therefore,$n^3–n=n(n–1)(n+1)$ is divisible by $6$.