Question

Sole this hard question

Answer 1

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.

 

Answer 2

let us consider x as the square of any positive integer.
let b=3  therefore r= 0, 1, 2
using euclid's division lemma, we get:
when r=0
x=(3q+0)²
x=(3q)²
x=9q²
x=3(3q²)
x=3m  ......................... where m=3q²


when r=1
x=(3q+1)²             using the identity (a+b)²= a²+2ab+b²
x=(3q)²+2×3q×1+1²
x=9q²+6q+1
x=3(3q²+2q)+1
x=3m+1     .......................where m= 3q²+2q