use euclid's division lemma to show that the square of any positive intiger is either of the form *3m* or *3m+1* for some integer *m*?

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

a^{2} = 9q^{2}

^{ }= 3 x ( 3q^{2})

= 3m (where m = 3q^{2})

Case II - a = 3q +1

a^{2} = ( 3q +1 )^{2}

= 9q^{2 }+ 6q +1

= 3 (3q^{2} +2q ) + 1

= 3m +1 (where m = 3q^{2} + 2q )

Case III - a = 3q + 2

a^{2} = (3q +2 )^{2}

= 9q^{2 }+ 12q + 4

= 9q^{2} +12q + 3 + 1

= 3 ( 3q^{2 }+ 4q + 1 ) + 1

= 3m + 1 where m = 3q^{2} + 4q + 1)

From all the above cases it is clear that square of any positive integer ( as in this case a^{2} ) is either of the form 3m or 3m +1.

- Sumit

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