Question

using euclid division lemma to show that square of any positive integer is always equal to the 3m,3m+1

Answer 1

Hey

let ' a' be any positive integer and b = 3.

we know, a = bq + r , 0 < r< b.

now, a = 3q + r , 0
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.

Hope this helps you ☺☺

Answer 2

hello ,
Here is Ur answer .★
Hope this helps you.::::::::::::::::::::::::::::::::::::::::::::::

Sol .

Let A be any position integer and b= 3.

Than a = 3q+r for some Integer q ≥ 0
and r = 0, 1 , 2 because 0 ≤ r < 3

Therefore, a = 3q or 3q+1 or 3q +2 or,
${a}^{2} = ( {3q)}^{2} or \: {(3q + 1)}^{2}or \: {(3 q + 2)}^{2} \\ \\ {a}^{2} = {(9q)}^{2} or \: 9 {q + 6q} + 1 \: or \: \: \: 9q +12 + 4 \\ \\ = > 3 \times (3 {q}^{2} ) \: or \: 3(3 {q}^{2} + 2q) + 1 \: \: or \\ \: \: \: \: \: \: \: \: \: \: \: \: \: 3(3 {q}^{2} + 4q + 1) + 1 \\ \\ = > 3k1 \: \: or \: 3k2 + 1 \: \: or \: 3k3 + 1$
where, k1 , k2 , or k3 are some positive integer.

Hence, it can be said that the square of any positive integer is either of the form 3m or 3m+1.

<<<☺>>>