Let, n be any positive integer. And since any positive integer can be of the form 6q, or 6q+1, or 6q+2, or 6q+3, or 6q+4, or 6q+5. (From Euclid’s division lemma for b= 6)
We have n3 – n = n(n2-1)= (n-1)n(n+1),
For n= 6q,
⇒ (n-1)n(n+1)= (6q-1)(6q)(6q+1)
⇒ (n-1)n(n+1)= 6[(6q-1)q(6q+1)]
⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= (6q-1)q(6q+1)]
For n= 6q+1,
⇒ (n-1)n(n+1)= (6q)(6q+1)(6q+2)
⇒ (n-1)n(n+1)= 6[q(6q+1)(6q+2)]
⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= q(6q+1)(6q+2)]
For n= 6q+2,
⇒ (n-1)n(n+1)= (6q+1)(6q+2)(6q+3)
⇒ (n-1)n(n+1)= 6[(6q+1)(3q+1)(2q+1)]
⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= (6q+1)(3q+1)(2q+1)]
For n= 6q+3,
⇒ (n-1)n(n+1)= (6q+2)(6q+3)(6q+4)
⇒ (n-1)n(n+1)= 6[(3q+1)(2q+1)(6q+4)]
⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= (3q+1)(2q+1)(6q+4)]
For n= 6q+4,
⇒ (n-1)n(n+1)= (6q+3)(6q+4)(6q+5)
⇒ (n-1)n(n+1)= 6[(2q+1)(3q+2)(6q+5)]
⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= (2q+1)(3q+2)(6q+5)]
For n= 6q+5,
⇒ (n-1)n(n+1)= (6q+4)(6q+5)(6q+6)
⇒ (n-1)n(n+1)= 6[(6q+4)(6q+5)(q+1)]
⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= (6q+4)(6q+5)(q+1)]
Hence, for any positive integer n, n3 – n is divisible by 6.