# Prove that for any positive interger n, n^3 - n is divisible by 6.

there are two methods to solve the problem which are discussed below.

Method 1:

Let us consider

a = n3 – n

a = n (n2 – 1)

a = n (n + 1)(n – 1)

Assumtions:

1. Out of three (n – 1), n, (n + 1) one must be even, so a is divisible by 2.

2. (n – 1) , n, (n + 1) are consecutive integers thus as proved a must be divisible by 3.

From (1) and (2) a must be divisible by 2 × 3 = 6

Thus, n³ – n is divisible by 6 for any positive integer n.

Method 2:

When a number is divided by 3, the possible remainders are 0 or 1 or 2.

∴ n = 3p or 3p + 1 or 3p + 2, where r is some integer.

Case 1: Consider n = 3p

Then n is divisible by 3.

Case 2: Consider n = 3p + 1

Then n – 1 = 3p + 1 –1

⇒ n -1 = 3p is divisible by 3.

Case 3: Consider n = 3p + 2

Then n + 1 = 3p + 2 + 1

⇒ n+1 = 3p + 3

⇒ n+1  = 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.

⇒ n (n – 1) (n + 1) is divisible by 3.

Similarly, when a number is divided by 2, the possible remainders are 0 or 1.

∴ n = 2q or 2q + 1, where q is some integer.

Case 1: Consider n = 2q

Then n is divisible by 2.

Case 2: Consider n = 2q + 1

Then n–1 = 2q + 1 – 1

n – 1 = 2q is divisible by 2 and

n + 1 = 2q + 1 + 1

n +1 = 2q + 2

n+1= 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.

∴ n (n – 1) (n + 1) is divisible by 2.

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

Therefore, as per the divisibility rule of 6, the given number is divisible by six.

n3 – n = n (n – 1) (n + 1) is divisible by 6.