Question

Prove that the square of any positive integer is of the form 3m or, 3m + 1 but not of the form 3m + 2.

Answer 1

SOLUTION :  

Since positive integer n is of the form of 3q , 3q + 1 and 3q + 2

Case : 1

If n = 3q,then

n² = (3q)²

[On squaring both sides]

n² = 9q²

n²= 3 (3q)²

n² = 3m , where m = 3q²

Case : 2

If n = 3q + 1,then

n² = (3q + 1)²

[On squaring both sides]

n² = (3q)² + 6q + 1²

[(a+b)² = a² + b² + 2ab]

n² = 9q² + 6q + 1

n² = 3q (3q + 2) + 1

n² = 3m +1 , where m = q(3q + 2)

Case : 3

If n = 3q + 2, then

n² = (3q + 2)²

[On squaring both sides]

n² = (3q)² + 12q + 4

[(a+b)² = a² + b² + 2ab]

n² = 9q² + 12q + 4

n² = 3 (3q² + 4q + 1) +1

n² = 3m + 1 , where q = 3q² + 4q + 1

Hence, n²  is of the form 3m, 3m + 1 but not of the form 3m +2.

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

hello mate
here is your solution

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)

I hope helps you