Question

show that the square of any two positive integers is of the form 3m or 3m+1 for some integer m

Answer 1

HOLA

====================

Let n be an arbitary integer

On diving n by 3 we get quotient M and remainder R

So we have , ( 3m ) , ( 3m + 1 )

On squaring we have

$( \: 3m \: ) {}^{2} = \: 9 \: ( \: clearly \: postive \: ) \\ \\ ( \: 3m \: + \: 1 \: ) {}^{2} = \: 9m \: + \: 1 \: ( \: clearly \: postive \: )$
So we conclude that any of the integer of form ( 3m ) , ( 3m + 1 ) is a positive integer for some integer M

========================

HOPE U UNDERSTAND ❤❤

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 .