Question

for any positive integer n show that n cube minus n is divisible by 6

Answer 1

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.

There is an alternate method too:

By the mathematical induction we have to prove this.
⇒  Given that S(n) = n3-n divisible by 6.
Let n =1 then we get '0'
which is divisible by 6.
  ∴ S(1) is true.
Let us assume that n = k
 S(k) = k3- k
which is divisible by 6.
∴ S(k) is true.
∴ (k3-k) / 6 = m ( integer )
  (k3-k)  = 6m
   k3= 6m +k --------→(1)
now we have to prove that n = k+1
⇒ (k+1)3 - (k+1)
⇒ (k3+3k2+3k+1) - (k+1)
subsitute equation (1) in above equation then
⇒ 6m +k+3k2+2k
⇒ 6m +3k2+k
⇒ 6m +3k(k+1)  ( ∴k(k+1) = 2p is an even number p is natural number)
⇒ 6m +3x2p
⇒ 6(m +p)
∴which is divisible by 6
s(k+1) is true.
By the mathematical induction it is true for n∈N.

Or,

n³ - n = n(n²-1) = n(n -1)(n + 1) is divided by 3 then possible reminder is 0, 1 and 2 [ ∵ if P = ab + r , then 0 ≤ r < a by Euclid lemma ] 

∴ Let n = 3r , 3r +1 , 3r + 2 , where r is an integer 

Case 1 :- when n = 3r 

Then, n³ - n is divisible by 3 [∵n³ - n = n(n-1)(n+1) = 3r(3r-1)(3r+1) , clearly shown it is divisible by 3 ] 

 

Case2 :- when n = 3r + 1 

e.g., n - 1 = 3r +1 - 1 = 3r 

Then, n³ - n = (3r + 1)(3r)(3r + 2) , it is divisible by 3 

 

Case 3:- when n = 3r - 1 

e.g., n + 1 = 3r - 1 + 1 = 3r 

Then, n³ - n = (3r -1)(3r -2)(3r) , it is divisible by 3 

 

From above explanation we observed n³ - n is divisible by 3 , where n is any positive integers 

 

Hope This Helps :)