Let s be any positive integer.
On dividing s by 6, let m be the quotient and r be the remainder.
By Euclid’s division lemma,
s = 6m + r, where 0 ≤ r ˂ 6
We can say that,
Any positive integer is of the form 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4, 6m + 5 for some positive integer n.
Case 1: When n = 6m
n3 – n = (6m)3 – 6m = 216 m3 – 6m = 6m(36m2 – 1) = 6q, where q = m(36m2 -1)
n3 – n is divisible by 6
Case 2: When n = 6m + 1
n3 – n = n(n2 – 1) = n (n – 1) (n + 1) = (6m + 1) (6m) (6m + 2) = 6m(6m + 1) (6m + 2)
= 6q,
where q = m(6m + 1) (6m + 2)
n3 – n is divisible by 6
Case 3: When n = 6m + 2
n3 – n = n (n – 1) (n + 1) = (6m + 2) (6m + 1) (6m + 3) = (6m + 1) (36 m2 + 30m + 6)
= 6[m (36m2 + 30m + 6)] + 6 (6m2 + 5m + 1)
= 6p + 6q, where p = m (36m2 + 30m + 6) and q = 6m2 + 5m + 1
n3 – n is divisible by 6
Case 4: When n = 6m + 3
n3 – n = (6m + 3)3 – (6m + 3) = (6m + 3) [(6m + 3)2 – 1] = 6m [6m + 3)2 – 1] + 3 [(6m + 3)2 – 1]
= 6 [m [(6m + 3)2 – 1] + 6 [18m2 + 18m + 4]
= 6p + 3q,
where p = m[(6m + 3)2 – 1]
q = 18m2 + 18m + 4
n3 – n is divisible by 6
Case 5: When n = 6m + 4
n3 – n = (6m + 4)3 – (6m + 4) = (6m + 4) [(6m + 4)2 – 1]
= 6m [(6m + 4)2 – 1] + 4 [(6m + 4)2 – 1]
= 6m [(6m + 4)2 – 1] + 4 [36m2 + 48m + 16 – 1]
= 6m [(6m + 4)2 – 1] + 12 [12m2 + 16m + 5]
= 6p + 6q,
where p = m [(6m + 4)2 – 1]
q = 2 (12 m2 + 16m + 5)
n3 – n is divisible by 6
Case 6: When n = 6m + 5
n3 – n = (6m + 5) [(6m + 5)2 – 1] = 6m [(6m + 5)2 – 1] + 5 [(6m + 5)2 – 1]
= 6p + 30q
= 6 (p + 5q),
where p = m [(6m + 5)2 – 1]
q = 6m2 + 10m + 4
n3 – n is divisible by 6
Therefore, n3 – n is divisible by 6, for any positive integer n.