Question

use euclid's division lemma to show that
the squre of any positive integer is either
of the from 3m or 3m+1 for some integer m? ​

Answer 1

Answer:

To prove:-

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

Solution:-

Let 'a' be any positive integer.

Using Euclid's division lemma;

a = 3q + r_____(1)

Where, r = 0, 1, 2.

Now, substituting given values of r,

in eq(1).

a = 3q

a = 3q + 1

a = 3q + 2

When, r = 0

a = 3q

Squaring both sides,

a² = (3q)²

a² = 9q²

= 3 × 3q² (splitting 9q²)

= 3m (taking 3q² as ‘m’)

When, r = 1

a = 3q + 1

Squaring both sides,

a² = (3q + 1)²

= 9q² + 1 + 6q

= 3(3q² + 2q) + 1 (taking 3q²+ 2q as ‘m’)

= 3m + 1

When, r = 2

a = 3q + 2

Squaring both sides,

a² = (3q + 2)²

= 9q² + 4 + 12q

= (9q² + 12q + 3) + 1

= 3(3q² + 4q + 1) +1

(taking 3q²+4q+1 as ‘m’)

= 3m + 1

Thus, square of any positive integer is of the form 3m or 3m+1 .

Answer 2

Case 1: Consider r = 0

Equation (1) becomes

a = 3b

On squaring both the side

a2 = (3b)2

a2 = 9b2

a2 = 3 × 3b2

a2 = 3m

Where m = 3b2

Case 2: Let r = 1

Equation (1) becomes

a = 3b + 1

Squaring on both the side we get

a2 = (3b + 1)2

a2 = (3b)2 + 1 + 2 × (3b) × 1

a2 = 9b2 + 6b + 1

a2 = 3(3b2 + 2b) + 1

a2 = 3m + 1

Where m = 3b2 + 2b

Case 3: Let r = 2

Equation (1) becomes

a = 3b + 2

Squaring on both the sides we get

a2 = (3b + 2)2

a2 = 9b2 + 4 + (2 × 3b × 2)

a2 = 9b2 + 12b + 3 + 1

a2 = 3(3b2 + 4b + 1) + 1

a2 = 3m + 1

where m = 3b2 + 4b + 1

∴ square of any positive integer is of the form 3m or 3m+1.

Hence proved.