Question

Prove that n square - n is divisible by 2 and 4 for some positive integer

Answer 1

Hi Mate !!


Here's ur ans of above question !

Let n be any positive integer Which when divided by 4 gives q as quotient and r as remainder.

According to Euclid's division lemma

a = bq + r

Where , 0 ≤ r < b

So , n = 4q + r
r = 0 , 1 , 2 , 3

n = 4q

n = 4q + 1

n = 4q + 2

n = 4q + 3

__________

• Case - 1

n = 4q

n² - n = ( 4q )² - 4q

= 16q² - 4q

= 4 ( 4q² - q )

= 4m [ Where m = 4q² - q ]

Therefore It is divisible by 4 and the no. which is divisible by 4 is divisible by 2 also.

• Case - 2

n = 4q + 1

n² - n = ( 4q + 1 )² - ( 4q + 1 )

{ Using identity ( a + b )² = a² + b² + 2ab }

= 16q² + 1 + 8q - 4q - 1

= 16q² + 4q

= 4 ( 4q² + q )

= 4m [ Where m = 4q² + q ]

Therefore It is divisible by both 2 and 4


• Case - 3

n = 4q + 2

n² - n = ( 4q + 2 )² - ( 4q + 2 )

= 16q² + 4 + 16q - 4q - 2

= 16q² + 12q + 2

= 4 ( 4q² + 3q ) + 2

= 4m + 2 [ Where m = 4q² + 3q ]

Therefore the no. is only divisible by 2


• Case - 4

n = 4q + 3

n² - n = ( 4q + 3 )² - ( 4q + 3 )

= 16q² + 9 + 24q - 4q - 3

= 16q² + 20q + 4 + 2

= 4 ( 4q² + 5q + 1 ) + 2

= 4m + 2 [ Where m = 4q² + 5q + 1 ]

Therefore It is divisible by 2 only.