Question

Prove that For any positive integer n^3-n is always divisible by 6.

Answer 1

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n^3 - n = n (n2 - 1) = n (n - 1) (n + 1) 

When a number is divided by 3 remainder is either 0,1,2

n = 3p , 3p + 1, 3p + 2 p is some integer

n = 3p 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 one of the numbers among n, n – 1 and n + 1 is always divisible by 3.

n (n – 1) (n + 1) is divisible by 3.
 
when a number is divided 2, the remainder obtained is 0 or 1.

n = 2q or 2q + 1
q is integer

n = 2q 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.

one of the numbers among n, n – 1 and n + 1 is always divisible by 2.

n (n – 1) (n + 1) is divisible by 2.

n (n – 1) (n + 1) is divisible by 2 and 3.

n (n-1) (n+1) = n^3 - n is divisible by 6

( number divisible by both 2 and 3 then it is divisible by 6)