Question

Show that the square of any positive integer is of the form 3m, 3m+1 for some integer 'm'.

Answer 1

Answer:

Let a be any positive odd integer and b =3.

Step-by-step explanation:

now using euclid division lemma a=bq+r , a=3q+r here r is remainder and it can be 0,1,2 because o=r<b and b is 3

so, all the possible euations are 3q, 3q+1, 3q+2

square of 3q                            = 9q2                             =         3m

(3q+1)2 = 9q2 + 1 + 6q2           = 3(3q2 + 2q2) +1          =         3m+1

(3q+2)2 = 9q2 + 4 + 12q2        = 3(3q2 + 4q2) +1          =        3m+1

hence proved

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.

Hence, it is solved .