Question
For any positive integer $n$, prove that $n^3-n$ is divisible by $6$ .
✓
Answer
$n^3-n=n\left(n^2-1\right)=n(n-1)(n+1)$
Whenever a number is divided by 3 , the remainder obtained is either $0$ or $1$ or $2$ .
$\therefore n=3 p$ or $3 p+1$ or $3 p+2$, where $p$ is some integer.
If $n =3 p$, then n is divisible by $3$ .
If $n=3 p+1$, then $n-1=3 p+1-1=3 p$ is divisible by $3$ .
If $n=3 p+2$, then $n+1=3 p+2+1=3 p+3=3(p+1)$ is divisible by $3$ .
So, we can say that one of the numbers among $n, n-1$ and $n+1$ is always divisible by $3$ .
$\Rightarrow n ( n -1)( n +1)$ is divisible by 3 .
Similarly, whenever a number is divided $2$ , the remainder obtained is $0$ or $1$ .
$\therefore n=2 q$ or $2 q+1$, where $q$ is some integer.
If $n=2 q$, then $n$ is divisible by $2$ .
If $n=2 q+1$, then $n-1=2 q+1-1=2 q$ is divisible by $2$ and $n+1=2 q+1+1=2 q+2=2(q+1)$ is divisible by
$2$.
So, we can say that one of the numbers among $n , n -1$ and $n +1$ is always divisible by $2$ .
$\Rightarrow n ( n -1)( n +1)$ is divisible by $2$ .
Since, $n(n-1)(n+1)$ is divisible by $2$ and $3$ .
$\therefore n(n-1)(n+1)=n^3-n$ is divisible by $6$ . (If a number is divisible by both $2$ and $3$ , then it is divisible by $6$ )
Whenever a number is divided by 3 , the remainder obtained is either $0$ or $1$ or $2$ .
$\therefore n=3 p$ or $3 p+1$ or $3 p+2$, where $p$ is some integer.
If $n =3 p$, then n is divisible by $3$ .
If $n=3 p+1$, then $n-1=3 p+1-1=3 p$ is divisible by $3$ .
If $n=3 p+2$, then $n+1=3 p+2+1=3 p+3=3(p+1)$ is divisible by $3$ .
So, we can say that one of the numbers among $n, n-1$ and $n+1$ is always divisible by $3$ .
$\Rightarrow n ( n -1)( n +1)$ is divisible by 3 .
Similarly, whenever a number is divided $2$ , the remainder obtained is $0$ or $1$ .
$\therefore n=2 q$ or $2 q+1$, where $q$ is some integer.
If $n=2 q$, then $n$ is divisible by $2$ .
If $n=2 q+1$, then $n-1=2 q+1-1=2 q$ is divisible by $2$ and $n+1=2 q+1+1=2 q+2=2(q+1)$ is divisible by
$2$.
So, we can say that one of the numbers among $n , n -1$ and $n +1$ is always divisible by $2$ .
$\Rightarrow n ( n -1)( n +1)$ is divisible by $2$ .
Since, $n(n-1)(n+1)$ is divisible by $2$ and $3$ .
$\therefore n(n-1)(n+1)=n^3-n$ is divisible by $6$ . (If a number is divisible by both $2$ and $3$ , then it is divisible by $6$ )
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