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

By BYJU'S Exam Prep

Updated on: September 25th, 2023

Steps to Prove that for any positive integer n, n3 – n is divisible by 6.

Step I: To prove the divisibility by 3.

The potential remainders after dividing an integer by three are 0 or 1 or 2.

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

Case 1: Consider n = 3p

If so, n can be divided 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

On simplifying we get

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.

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Step II: To prove the divisibility by 2.

Similar to this, there are only two potential remainders when an integer is divided by two: 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.

In other words, one of the numbers between n, n – 1, and n + 1 is always divisible by two.

n (n – 1) (n + 1) is divisible by 2.

Step III: To prove the divisibility by 6

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

As a result, the supplied number can be divided by 6 according to the law of divisibility of 6.

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

Summary:

It is proved that for any positive integer n, n3 – n is divisible by 6.