Question

Use euclids division leema to show that the squars of any positive integer is either of form 3m or 3m +1 for some k teger m

Answer 1

3
Let a be any positive integer and b = 3.
Then a = 3q + r for some integer q ≥ 0
And r = 0, 1, 2 because 0 ≤ r < 3
Therefore, a = 3q or 3q + 1 or 3q + 2 Or,
푎2= (3푟)2 표푠 (3푟 + 1)2 표푠 (3푟 + 2)2
= (3푟)2 표푠 9푟2+ 6푟 + 1 표푠 9푟2+ 12푟 + 4
= 3 × (3푟2) 표푠 3 × (3푟2+ 2푟) + 1 표푠 3 × (3푟2+ 4푟 + 1) + 1
= 3푘1 표푠 3푘2+ 1 표푠 3푘3+ 1
Where k1, k2, and k3 are some positive integers
Hence, it can be said that the square of any positive integer is either of
the form 3m or 3m + 1.
I have written it by myself please mark it as the brainllist if it will help you

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.