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

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

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n 3 – n = n(n2 – 1) = n(n – 1)(n + 1)

For a number to be divisible by 6, it should be divisible by 2 and 3 both,

Divisibility by 3:

n – 1, n and n + 1 are three consecutive whole numbers.

By Euclid’s division lemma

n + 1 = 3q + r, for some integer k and r < 3

As,

r < 3 possible values of ‘r’ are 0, 1 and 2.

If r = 0

n + 1 = 3q

⇒ n + 1 is divisible by 3

⇒ n(n – 1)(n + 1) is divisible by 3

⇒ (n3 – n) is divisible by 3

If r = 1

⇒ n + 1 = 3q + 1

⇒ n = 3q

⇒ n is divisible by 3

⇒ n(n – 1)(n + 1) is divisible by 3

⇒ (n3 – n) is divisible by 3

If r = 2

⇒ n + 1 = 3q + 2

⇒ n + 1 – 2 = 3q

⇒ n – 1 = 3q

⇒ n - 1 is divisible by 3

⇒ n(n – 1)(n + 1) is divisible by 3

⇒ (n3 – n) is divisible by 3

Divisibility by 2:

If n is even

Clearly,

n(n – 1)(n + 1) is divisible by 2

If n is odd

⇒ n + 1 is even

⇒ n + 1 is divisible by 2

⇒ n(n – 1)(n + 1) is divisible by 2

Hence,

For any value of n, n3 - n is divisible by 2 and 3 both,

therefore n3 – n is divisible by 6.