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

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Answer 2

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Step-by-step explanation:

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;

x^2 = (3q)^2 = 9q^2 = 3.3q^2

Let 3q^2 = m

Therefore,

x^2 = 3m ………………….(1)

x^2 = (3q + 1)^2

= (3q)^2 + 12 + 2 × 3q × 1

= 9q^2 + 1 + 6q

= 3(3q^2 + 2q) + 1

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

x^2 = 3m + 1 ……………………………. (2)

x^2 = (3q + 2^)2

= (3q)^2 + 22 + 2 × 3q × 2

= 9q^2 + 4 + 12q

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

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

x^2 = 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.