Question

use Euclid division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 for some integer m​

Answer 1

Let 'a' be any positive integer.

On dividing it by 3 , let 'q' be the quotient and 'r' be the remainder.

Such that ,

a = 3q + r , where r = 0 ,1 , 2

When, r = 0

∴ a = 3q

When, r = 1

∴ a = 3q + 1

When, r = 2

∴ a = 3q + 2

When , a = 3q

On squaring both the sides,

${a}^{2} = 9 {q}^{2} \\ {a}^{2} = 3 \times (3 {q}^{2} ) \\ {a}^{2} = 3 \\ where \: m = 3 {q}^{2}$

When, a = 3q + 1

On squaring both the sides ,

${a}^{2} = (3q + 1)^{2} \\ {a}^{2} = 9 {q}^{2} + 2 \times 3q \times 1 + {1}^{2} \\ {a}^{2} = 9 {q}^{2} + 6q + 1 \\ {a}^{2} = 3(3 {q}^{2} + 2q) + 1 \\ {a}^{2} = 3m + 1 \\ where \: m \: = 3 {q}^{2} + 2q$

When, a = 3q + 2

On squaring both the sides,

${a}^{2} = (3q + 2)^{2} \\ {a}^{2} = 3 {q}^{2} + 2 \times 3q \times 2 + {2}^{2} \\ {a}^{2} = 9 {q}^{2} + 12q + 4 \\ {a}^{2} = (9 {q}^{2} + 12q + 3) + 1 \\ {a}^{2} = 3(3 {q}^{2} + 4q + 1) + 1 \\ {a}^{2} = 3m + 1 \\ where \: m \: = 3 {q}^{2} + 4q + 1$

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

Answer 2

Use Euclid division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 for some integer m.

Let a be any positive integer and b = 3.

=) a = 3q + r, r = 0 or 1 or 2.

(By Euclid's lemma)

=) a = 3q or 3q + 1 or 3q + 2 for positive integer q.

1st case,

If a = 3q :

=) a² = (3q)²

= 9q²

= 3(3q²)

= 3m, where m= 3q².

2nd case,

If a = 3q+1,

=) a² = (3q+1)²

= (3q)² + 2(3q)(1) + 1²

= 9q² + 6q + 1

= 3(3q² + 2q) + 1

= 3m + 1, where m = 3q² + 2q.

3rd case,

If a = 3q+2:

=) a² = (3q+2)²

= (3q)² + 2(3q)(2) + 2²

= 9q² + 12q + 4

= 9q² + 12q + 3 + 1

= 3(3q² + 4q + 1) + 1

= 3m + 1, where m = 3q² + 4q + 1.

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