Question

use euclid's division Lemma to show that square of any positive integer is either in the form of 3 or 3m + 1

Answer 1

let a be any of positive integer which is divisible by 3 for some integer q and r
then,
Using E.D.L.
a=3q+r where 0<=r<q
so a=3q or 3q+1 or 3q+2
when a =3q
squaring both side
a^2=(3q)^2
a^2=9q^2
a^2=3(3q^2)
a^2=3m where m=3q^2
when a=3q+1
squaring both side
a^2=(3q+1)^2
a^2=9q^2+1+6q
a^2=3(3q^2+2q)+1
a^2=3m+1 where m=3q^2+2q
when a=3q+2
squaring both side
a^2=(3q+2)^2
a^2=9q^2+12q+4
a^2=9q^2+12q+3+1
a^2=3(3q^2+4q+1)+1
a^2=3m+1 where m =3q^2+4q+1

Therefore square of any positive integer is divisible by 3m or 3m+1.

Answer 2

Step-by-step explanation:

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.