For any positive integer n, prove that n^3 – n is divisible by 6.

Let a = n^3 - n ⇒ a = n.(n^2 - 1)
⇒ a = n.(n-1).(n+1)  [∵(a^2 - b^2) = (a-b)(a+b)]
⇒ a = (n-1).n.(n+1)  ...(i)
We know that,
1. If a number is completely divisible by 2 and 3, then it is also divisible by 6.
2. If the sum of digits of any number is divisible by 3, then it is also divisible by 3.
3. If one of the factor of any number is an even number,then it is also divisible by 2.
∴ a = (n-1).n.(n+1)  [from Eq.(i)]
Now,sum of the digits = n-1+n+n+1 = 3n
= multiple of 3,where n is any positive integer,
and (n-1)-n-(n+1) will always be even, as one out of (n-1) or n or (n+1) must of even.
Since, conditions II and III is completely satify the Eq.(i).
Hence, by condition I the number n^3 - n is always divisible by 6, where n is any positive integer.
Hence proved.

Solution:
n³ – n = n (n² – 1) = n (n – 1) (n + 1)
Therefore, n³ – n is product of three consecutive positive integers, where n is any positive integer.
Since one out of every two consecutive integers is divisible by 2.

Therefore, The product n³ – n is divisible by 2.
Since one out of every three consecutive integers is divisible by 3.
Therefore, The product n³ – n is divisible by 3.
Any number which is divisible by 2 and 3 is also divisible by 6.
Hence, The product n³ – n is divisible by 6.

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.