Question

For any positive integer n, prove that n3-n is divisible by 6.

Answer 1

n³ - n = n(n²-1) = n(n -1)(n + 1) is divided by 3 then possible reminder is 0, 1 and 2 [ ∵ if P = ab + r , then 0 ≤ r < a by Euclid lemma ]
∴ Let n = 3r , 3r +1 , 3r + 2 , where r is an integer
Case 1 :- when n = 3r
Then, n³ - n is divisible by 3 [∵n³ - n = n(n-1)(n+1) = 3r(3r-1)(3r+1) , clearly shown it is divisible by 3 ]

Case2 :- when n = 3r + 1
e.g., n - 1 = 3r +1 - 1 = 3r
Then, n³ - n = (3r + 1)(3r)(3r + 2) , it is divisible by 3

Case 3:- when n = 3r - 1
e.g., n + 1 = 3r - 1 + 1 = 3r
Then, n³ - n = (3r -1)(3r -2)(3r) , it is divisible by 3

From above explanation we observed n³ - n is divisible by 3 , where n is any positive integers

Answer 2

Let :-

a = n³ - n

= n(n² - 1)

= n(n - 1)(n + 1)

= (n - 1)n (n + 1)

1) Now out of three (n - 1),n and (n + 1) one must be even so a is divisible by 2

2) Also (n - 1),n and (n + 1) are three consecutive integers thus as proved a must be divisible by 3

From (1) and (2)

a must be divisible by 2 × 3 = 6

Hence n³ - n is divisible by 6 for any positive integer n