Question

Use Euclid’s 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 x be any positive integer and y = 3.

By Euclid’s division algorithm;

x =3q + r (for some integer q ≥ 0 and r = 0, 1, 2 as r ≥ 0 and r < 3)

Therefore,

x = 3q, 3q + 1 and 3q + 2

As per the given question, if we take the square on both the sides, we get;

x2 = (3q)2 = 9q2 = 3.3q2

Let 3q2 = m

Therefore,

x2 = 3m ………………….(1)

x2 = (3q + 1)2

= (3q)2 + 12 + 2 × 3q × 1

= 9q2 + 1 + 6q

= 3(3q2 + 2q) + 1

Substitute, 3q2+2q = m, to get,

x2 = 3m + 1 ……………………………. (2)

x2 = (3q + 2)2

= (3q)2 + 22 + 2 × 3q × 2

= 9q2 + 4 + 12q

= 3(3q2 + 4q + 1) + 1

Again, substitute, 3q2 + 4q + 1 = m, to get,

x2 = 3m + 1…………………………… (3)

Hence, from eq. 1, 2 and 3, we conclude that, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

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