Question

use Euclid's division lemma to show that the square of any positive integer is either of the form 3mor 3m+1 for some integer m.​

Answer 1

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.

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.