Question

How do I answer this question?

Answer 1

let a be any arbitrary integer and

b = 3.

we know,(Euclid Algorithm)

a = bq + r , 0 <  r< b.


Now,


a = 3q + r , 0 < r < 3.

The possibility of remainder = 0,1 or 2

=========================
$\bold{ \huge{case - 1}}$


a = 3q

a2 = 9q2

  3 . ( 3q2)

  3m (here, m = 3q²)




$\bold{ \mathfrak{Case 2 - - - - }}$



a = 3q +1

a2 = ( 3q +1 )2

  9q2 + 6q +1

  3 (3q2 +2q ) + 1

  3m +1 (where m = 3q2 + 2q )



$\bold{ \frak{Case III - }}$




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)


Thus,


3m and 3m+1 are Square of any positive Integer


Thanks!!

Answer 2

Step-by-step explanation:

Let a be any +ve integer and b will take the form 3q and 3q+1

Case i) when a=3q

Squaring on both sides

a2= 3q2

= 3q×3q

=3(3q2) [3q square]

=3m where m = 3q2

Case ii) when a= 3q+1

a2=(3q+1)2 [ 3q+1 the whole square]

=3q2+2×3q×1+(1)2 [3q the whole square and 1 the whole square]

=9q2 +6q+1

=3(3q2+2q) +1=3m+1

Where m=3q2+2q

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