Question

Prove that $6$ has no common factor with n then $n^2-1$ is divisible by $6$.

Answer 1

Solution -

If n and 6 has no common factors

that means n and 2 has no common factor (n is odd)

and n and 3 has no common factor also

$n^2-1=(n+1)(n-1)$ If n is odd then (n+1) & (n-1) are both

even so divisible by 2.

Every $3^{rd}$ numbers in natural numbers

is divisible by 3, so are of (n-1) as

(n+1) must be divisible by 3.

so, $n^2-1$ is divisible by 6.