Question

using Euclid division algorithm show that square of any positive integer is in the form of 3M or 3M + 1 for some integer m

Answer 1

Let the positive integer be 'a'
By Euclids division algorithm
a=bq+r eq1 ;0 </=r <b
substitute b=3 in 1
a=3q+r ;r=0,1,2
If r=0,a=3q+r
a square =(3q)square
a square =9q square
=3 (3qsquare )
a=3m;m=3qsquare eq2
If r=1,a=3q+1
a square =(3q+1)square
a square =(3q) square+2 (3q)(1)+1square
=9q square +6q+1
=3 (3q square +2q)+1
a square =3m+1. eq3
If r=2,a=3q+2
a square =(3q+2)square
=(3q)square+2 (3q(2)+(2)square
=9q square +12q+3+1
a=3m+1. eq4
From equation 2, 3and4 we can conclude that
square of any positive integer is either of the form 3M OR 3M+1
THANK YOU
HOPE IT'LL HELP YOU

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 .