Question

For any Positive integer $n$ , prove that $n^3-n$ divisible by $6$.

Answer 1

Hint: We have to check whether $n^3-n$ is divisible by $6$ or not. For this firstly we find the factors of which are $2$ and $3$ . To check whether a number is divisible by $6$ we check whether it is divisible by $2$ and $3$ or not. We check it by applying a division algorithm. It will give us a remainder and we check the division by $2$ and $3$ for each value of remainder.

Complete step-by-step answer:
We have to check whether $n^3-n$ is divisible by $6$ or not where $n$ is a positive integer.
The factors of $6$ are $2$ and $3$. So if the number is divisible by $2$ and $3$ then it will also be divisible by $6$.
Here $n^3-n=n(n^2-1)\Rightarrow n(n+1)(n-1)$
When we divide by $3$ we a number $q$ such that
$\text{number = 3q + r}$ where $r$ is remainder and $r$ takes value $0,1\text{ and }2$.
$n=3p\text{ or }n=3q+1\text{ or }n=3q\text{ + 2}$
If $n=3p$ then $n$ will be divided by $3\Rightarrow n(n-1)(n+1)=n^3-n$ is also divided by $3$.
If $n=3p\text{ + 1}$ then $n-1=3p+1-1\Rightarrow 3p$
So $(n-1)$ is divided by $3$
$\Rightarrow n(n-1)(n+1)=n^3-n$ Is divided by $3$
If $n=3q+2$ then $n+1=n$
$3q+2+1=3q+3$
$=3(q+1)$
So $(n+1)$ is divisible by $3$
$\Rightarrow (n+1)(n-1)=n^3-n$ is divisible by $3$
So number $n^3-n$ is divisible by in all $3$ cases
If number is divisible by $2$ then there exist $q$ such that
$number=2q+r$ where $r$ is remainder $r$ takes value $r=0$ and $1$ .
$\Rightarrow (n+1)(n-1)=n^3-n$ divided by $2$
So in both cases $n^3-n$ is divided by $2$
$\Rightarrow n^3-n$ is divided by both $2\text{ and }3$
So $n^3-n$ is divided by $6$

Note: Division algorithm states that for any positive integer $a\text{ and }b$ there exist the integers $q\text{ and }r$ such that $a=bq+r\text{ where }0⩽r<b$
The number $q$ is called quotient and $r$ is called remainder.