Prove that if 6 has no common factor with n, n² — 1 is divisible by 6.

Given here, there is no common factor between 6 and n.

Therefore, 6 and n are two distinct natural numbers.

We know that 6 has 2 and 3 as prime factors.

n can be written as n = 2k + 1 for all

Therefore, we can see that is divisible by 2.

Similarly, n can be written as n = 3k + 1 for all

Therefore, we can see that is divisible by 3.

Similarly, n can also be written as n = 3k–1 for all

Therefore, we can see that is divisible by 3.

So, n² — 1 is divisible by 2 and 3 both.

Since, 2 and 3 are prime numbers.

Therefore, n² — 1 is divisible by 2 × 3 = 6.