Use Euclid's division lemma to show that the square of any positive in

1.	Use Euclid's division lemma to show that the square of any positive integer is either ofthe form 3m or 3m + 1 for some integer m.
Answer» 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) 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. Like my answer if you find it useful!

Use Euclid's division lemma to show that the square of any positive integer is either ofthe form 3m or 3m + 1 for some integer m.

Answer»

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)

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.

Like my answer if you find it useful!