Question

Q. Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m. [Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]​

Answer 1

$\huge \underline{ \underline \mathfrak{ solution - }}$

Let n be an arbitrary positive integer. On dividing n by 3, let q be the quotient and r be the remainder.

Then by Euclid’s division lemma,

n =3q+r, where 0<r<3 i.e r =0,1, 2.

$\therefore {n}^{2} = 9 {q}^{2} + {r}^{2} + 6qr.......(1)$

Case 1- when r =0

Putting r=0 in (1),

${n}^{2} = 9 {q}^{2} = 3( {3q}^{2} ) = 3m$

Where m=3q² is an integer.

Case 2 - when r=1

Putting r=1 in (1),

${n}^{2} = ( {9q}^{2} + 1 + 6q) = 3( {3q}^{2} + 2q) + 1$

=3m+1, where m =(3q²+2q)is an integer.

Case 3 - when r =2

Putting r=2 in (1),

${n}^{2} = ( {9q}^{2} + 4 + 12q) = 3( {3q}^{2} + 4q + 1) + 1$

=3m+1, where m=(3q²+4q+1)is an integer.

Hence the square of any positive integer is of the form 3m or (3m+1)for some integer m.

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.