## for any positive integer n, prove that n cube minus n is divisible by 6​

Question

for any positive integer n, prove that n cube minus n is divisible by 6​

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3 months 2021-07-25T18:16:15+00:00 1 Answers 0 views 0

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.

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