Question

Prove that if 6 has no common factor with n, n²–1 is divisible by 6.

Answer 1

6 has no common factor with n , it means n is not divisible by 6 .
we know 2 and 3 are prime factors of 6.
so, n is not divisible by 2 and also not divisible by 3.

case 1 :- if n is not divisible by 2. it means n must be an odd integer.
let n = 2k + 1
so, n² - 1 = (2k + 1)² - 1
= 4k² + 4k + 1 - 1
= 4k² + 4k = 4k(k + 1)
hence, n² - 1 is divisible by 2.

case 2 :- if n is not divisible by 3 .
let n = 3k + 1
so, n² - 1 = (3k + 1)² - 1
= 9k² + 6k + 1 - 1
= 3k(3k + 2)
hence, n² - 1 is divisible by 3.

from case 1 and case 2 it is clear that if n is not divisible by 2 , n² -1 is divisible by 2 and if n is not divisible by 3 , n²-1 is divisible by 3 . then, it is clear that if n is not divisible by 6 , n² - 1 is divisible by 6.

Answer 2

Hi ,

6 and n have no common factors .

n = 6m + 1 or n = 6m + 5 , m € z+

i ) n² - 1 = ( 6m + 1 )² - 1

= ( 6m +1 )( 6m + 1 - 1 )

= ( 6m + 1 ) 6m

Clearly it is divisible by 6.

ii ) n² - 1 = ( 6m + 5 )² - 1

= ( 6m + 5 + 1 )( 6m + 5 - 1 )

= 6( m + 1)( 6m + 4 )

Clearly it is divisible by 6 .

Therefore ,

6 and n have no common factor then

n²- 1 is divisible by 6.

I hope this helps you.

: )