Question

For any positive integers n, prove that $n^3-n$ is divisible by 6.

Answer 1

divisible by 3 implies that is also divisible by 6

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 , that is also divisible by 6 where n is any positive integers