Question

By using Euclid division lemma show that the square of any postive integer is either 3m or 3m+1

Answer 1

let a be any positive integer and b =3.By applying Euclid's division lemma there exist unique integers q and r such that a=3q + r where 0≤r<3
since 0 ≤ r < 3 the possible remainders are 0,1 & 2
in case I
a=3q
a²=(3q)²
a²=9q²
a²=3(3q²)
a²=3m where m = 3q²

In case II
a=3q+1
a²=(3q+1)²
a²=9q²+6q+1
a²=3(3q²+2q)+1
a²=3q+1,where m = 3q²+2q

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.