US22. show that the square of any positive integer is either of thefor

1.	US22. show that the square of any positive integer is either of theform zm or 3mt) for some integerm..for 69+1,
Answer» Let a be any odd positive integer. We apply the division lemma with a and b = 3 . Since 0 ≤ r < 3 , the possible remainders are 0 , 1 and 2 . That is , a can be 3q , or 3q + 1 , or 3q + 2 , where q is the quotient . Now , a² = ( 3q )² = 9q² Which can be written in the form = 3 ( 3q² ) = 3m , [ since m = 3q² ] It is divisible by 3 . Again , a² = ( 3q + 1 )² = 9q² + 6q + 1 = 3( 3q² + 2q ) + 1 = 3m + 1 [ since m = 3q² + 2q , 3( 3q² + 2q ) is divisible by 3 ] Lastly , a² = ( 3q + 2 )² = 9q² + 12q + 4 = ( 9q² + 12q + 3 ) + 1 = 3( 3q² + 4q + 1 ) + 1 Which can be written in the form 3m + 1 , since 3( 3q² + 4q + 1 ) is divisible by 3. Therefore , The square of any positive integer is either of the form 3m or 3m + 1 for some integer m .

US22. show that the square of any positive integer is either of theform zm or 3mt) for some integerm..for 69+1,

Answer»

Let a be any odd positive integer.

We apply the division lemma with

a and b = 3 .

Since 0 ≤ r < 3 ,

the possible remainders are 0 , 1

and 2 .

That is , a can be 3q , or 3q + 1 , or

3q + 2 , where q is the quotient .

Now ,

a² = ( 3q )² = 9q²

Which can be written in the form

= 3 ( 3q² )

= 3m , [ since m = 3q² ]

It is divisible by 3 .

Again ,

a² = ( 3q + 1 )²

= 9q² + 6q + 1

= 3( 3q² + 2q ) + 1

= 3m + 1 [ since m = 3q² + 2q ,

3( 3q² + 2q ) is divisible by 3 ]

Lastly ,

a² = ( 3q + 2 )²

= 9q² + 12q + 4

= ( 9q² + 12q + 3 ) + 1

= 3( 3q² + 4q + 1 ) + 1

Which can be written in the form

3m + 1 , since

3( 3q² + 4q + 1 ) is divisible by 3.

Therefore ,

The square of any positive integer

is either of the form 3m or 3m + 1

for some integer m .