Question

use euclid's division Lemma to show that the square of any positive integer is either of the form of 3m and 3m+ 1 for some integer m​

Answer 1

Let take a as any positive integer and b = 3.

Then using Euclid�s algorithm we get�a = 3q + r� here r is remainder and value of q is more than or equal to 0 �and r = 0, 1, 2 because 0 < r < b� and the value of b is 3 So our possible values will 3q+0 , 3q+1 and 3q+2

Now find the square of values�

Use the formula (a+b)² = a² + 2ab +b² to open the square bracket�

(3q)²� ���������� = 9q²�� if we divide by 3 we get no remainder �

we can write it as 3*(3q²) �so it is in form of 3m� here m = 3q²

(3q+1)² ������� = (3q)² + 2*3q*1� + 1² ���� �

=9q² + 6q +1 now divide by 3 we get 1 remainder

so we can write it as 3(3q² + 2q) +1 so we can write it in form of 3m+1 and value of m is 3q² + 2q� here

(3q+2)² ������� = (3q)² + 2*3q*2� + 2²�

=9q² + 12q +4 �now divide by 3 we get 1 remainder

so we can write it as 3(3q² + 4q +1) +1 so we can write it in form of 3m +1 and value of m will 3q² + 4q +1

Square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Answer 2

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.

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