Question

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

Answer 1

this is the solution

Answer 2

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

a^2 = 9q^2

 = 3 x ( 3q^2)

 = 3m (where m = 3q2)

Case II)  a = 3q +1

a^2 = ( 3q +1 )^2

 =  9q^2 + 6q +1

 = 3 (3q2 +2q ) + 1

 = 3m +1 (where m = 3q^2 + 2q )

Case III - a = 3q + 2

a^2 = (3q +2 )^2

   = 9q^2 + 12q + 4

   = 9q^2 +12q + 3 + 1

 = 3 (3q^2 + 4q + 1 ) + 1

 = 3m + 1 where m = 3q^2 + 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.