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 ${n^3} - n = n({n^2} - 1){\text{ }} \Rightarrow {\text{ }}n(n + 1)(n - 1)$
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 $3{\text{ }} \Rightarrow {\text{ }}n(n - 1)(n + 1){\text{ }} = {n^3} - n$ is also divided by $3$.
If $n = 3p{\text{ + 1}}$ then $n - 1 = 3p + 1 - 1{\text{ }} \Rightarrow {\text{ }}3p$
So $(n - 1)$ is divided by $3$
$ \Rightarrow {\text{ }}n(n - 1)(n + 1){\text{ }} = {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 {\text{ }}{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 $a = bq + r{\text{ where }}0 \leqslant r < b$
The number $q$ is called quotient and $r$ is called remainder.