Question

for any positive integer n, prove that n3-n is divisible by 6

Answer 1

Answer:

Step-by-step explanation:

n3 - n = n (n2 - 1) = n (n - 1) (n + 1)

Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.

∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.

If n = 3p, then n is divisible by 3.

If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.

If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.

So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.

⇒ n (n – 1) (n + 1) is divisible by 3.

Similarly, whenever a number is divided 2, the remainder obtained is 0 or 1.

∴ n = 2q or 2q + 1, where q is some integer.

If n = 2q, then n is divisible by 2.

If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.

So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.

⇒ n (n – 1) (n + 1) is divisible by 2.

Since, n (n – 1) (n + 1) is divisible by 2 and 3.

∴ n (n-1) (n+1) = n3 - n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6)

Answer 2

n^3 – n = n(n^2 –1)

= n(n + 1)(n – 1)

= (n – 1) n(n + 1)

= Product of three consecutive positive integers  Now, we have to show that the product of  three consecutive positive integers is divisible  by 6.

We know that any positive integer a is of the  form 3q, 3q + 1 or 3q + 2 for some integer q.

Let a, a + 1, a + 2 be any three consecutive  integers.

If a = 3q

a(a + 1)(a + 2) = 3q(3q + 1)(3q + 2)

= 3q (2r)

= 6qr, which is divisible by 6.

If a = 3q + 1

a(a + 1)(a + 2) = (3q + 1)(3q + 2)(3q + 3)

= (2r) (3)(q + 1)

= 6r(q + 1) which is divisible by 6

If a = 3q + 2

a(a + 1)(a + 2) = (3q + 2)(3q + 3)(3q + 4)  = multiple of 6 for every  q = 6r (say) which is divisible by 6.

Therefore the product of three consecutive  integers is divisible by 6.