Question

Prove that the square of any positive integer is from 3m or 3m +1 but not in the form of 3m+2

Answer 1

Hi ,

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 .

I hope this helps you.

: )

Answer 2

Step-by-step explanation:

let ' a' be any positive integer and b = 3.

we know, a = bq + r , 0 <  r< b.

now, a = 3q + r , 0<r < 3.

the possibilities of remainder = 0,1 or 2

Case I - a = 3q

a² = 9q² .

 = 3 x ( 3q²)

 = 3m (where m = 3q²)

Case II - a = 3q +1

a² = ( 3q +1 )²

 =  9q² + 6q +1

 = 3 (3q² +2q ) + 1

 = 3m +1 (where m = 3q² + 2q )

Case III - a = 3q + 2

a² = (3q +2 )²

   = 9q² + 12q + 4

   = 9q² +12q + 3 + 1

 = 3 (3q² + 4q + 1 ) + 1

 = 3m + 1 where m = 3q2 + 4q + 1)

From all the above cases it is clear that square of any positive integer ( as in this case a² ) is either of the form 3m or 3m +1.

Hence, it is solved .

THANKS

#BeBrainly .