Question

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

Answer 1

$n^3-n=(n^2-1)=n(n-1)(n+1)$.
Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.
$\therefore$ n=3p or 3p +1 or 3p +2, where p is some integer.
If n=3p, then n is divisible by 3.
If n=3p + 1, then n -1 = 3p +1-1=3p is divisible by 3.
If n = 3p +2, then n + 1=3p+2+1=3p+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 = 2q or 2q+1, where q is some integer.
If n=2q, then n is divisible by 2.

If n=2q + 1, then n-1 = 2q + 1 – 1 = 2q is divisible by 2 and n+1=2q+1+1=2q+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)