Question

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

Answer 1

let 'a' be any positive integer and 'b' = 3

by Euclid's division lemma :-

a = bq + r where 0 ≤ r < b

therefore r = 0, 1 or 2

when r = 0, a = 3q + 0 --------(i)

when r = 1, a = 3q + 1 --------(ii)

when r = 2, a = 3q + 2 --------(iii)

squaring both sides of each equation.

➡ a² = (3q)²

= 9q²

= 3(3q²)

= 3m ---------(iv)

➡ a² = (3q + 1)²

using identity (a + b)² = a² + 2ab + b²

= (3q)² + 2(3q)(1) + (1)²

= 9q² + 6q + 1

= 3(3q² + 2q) + 1

= 3m + 1 ---------(v)

➡ a² = (3q + 2)²

using identity (a + b)² = a² + 2ab + b²

= (3q)² + 2(3q)(2) + (2)²

= 9q² + 12q + 4

= 9q² + 12q + 3 + 1

= 3(3q² + 4q + 1) + 1

= 3m + 1 ---------(vi)

by equation (iv), (v) and (vi) we get that any positive number is either in the form of 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.