Question

use euclids division lemma to show that the squre of any postive integer is either of the form 3m or 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

Answer:

let a=3m , 3m+1

a²=(3m)²

  =3×3 m²

  = 3 (3m²)

  =3 k  where k is any integer and k= 3m²

now take  a= 3m+1

a²=(3m+1)²

   =(3m)²+2 (3m)(1)+1²

   = 3×3 m² +2×3m +1

   =3(3m²+2m)+1

   =3b+1 where b is any integer and b=3m²+2m

therefore the square of any positive integeris either of the form 3m or 3m+1 for some integer m.