Question

Show that n3-n is divisible by 8 if n is an odd positive integer

Answer 1

Given :  n³ - n     , f n is an odd positive integer

To find : n³ - n  is divisible by 8 if n is an odd positive integer

Step-by-step explanation:

n³ - n  is divisible by 8 if n is an odd positive integer

n³ - n

= n(n² - 1)

if n is an odd positive number

let say n = 2k + 1

=> (2k + 1)(  (2k + 1)² - 1)

= (2k + 1) ( 4k² + 1 + 4k - 1)

= (2k + 1)(4k²  + 4k)

= 4k(2k + 1)(k + 1)

= 4 k(k + 1) (2k + 1)

Case 1 : if k is odd then k + 1 is even

=> 4(k + 1) is Divisible by 8

hence n³ - n  is divisible by 8

Case 2 : if  k is even

then 4k is divisible by 8

hence n³ - n  is divisible by 8

From both its proved

that  n³ - n  is divisible by 8 if n is an odd positive integer

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