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

Let a be any positive integer

Let b = 3

Then a = 3q + r

a = 3q , a = 3q + 1 and a = 3q + 2

If a = 3q for some integer q

Then

a² => 9q² = 3(3q²)

a² = 3m (m = 3q² for some integer)

If a = 3q + 1 for some integer q

Then

a² => (3q + 1)² = 9q² + 6q + 1

= 3(3q² + 2q) + 1

= 3m + 1 (m = 3q² + 2q for some integer )

If a = 3q + 2 for some integer q

Then

a² => (3q + 2)² = 9q² + 12q + 3 + 1

= 3(3q² + 4q + 1) + 1

= 3m + 1 (m = 3q² - 4q + 1 for some integer)

Hence the square of any positive integer is of form 3m or 3m + 1 for some integer m

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.