# 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

a2 = 9q2

= 3 x ( 3q2)

= 3m (where m = 3q2)

Case II - a = 3q +1

a2 = ( 3q +1 )2

=  9q+ 6q +1

= 3 (3q2 +2q ) + 1

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

Case III - a = 3q + 2

a2 = (3q +2 )2

= 9q+ 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.

- Sumit

• 218
• -14
What are you looking for?