Question

Use Euclid division lemma to show that the square of any positive integer is of the from 3p,3p+1

Answer 1

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Answer 2

Let 'a' be any positive integer and b = 3.

we know, a = bq + r , 0 <  r< b.

now, a = 3q + r , 0<r < 3.

r = 0 or 1 or 2

So,

a = 3q or 3q + 1 or 3q + 2

Case I

a = 3q

a² = 9q²

  = 3( 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.