Question

use Euclids division lemma to show that ths square of any postive integer is either of the form 3m or 3m+1 for some integer m?

Answer 1

a= bq +r., 0-<r <b. let b=3 ,then r = 0,1,2.. ( 3q)^2= 9q^2= 3×3q^2=3m where m= 3q^2. now (3q+1)^2= 9q^2 + 6q +1=3(3q^2+2q)+1=3m+1 where m= 3q^2+2q.

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.