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

$\huge\underline\mathrm{Question}:-$

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.

$\huge\underline\mathrm{Solution}:-$

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 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 equations 1, 2 and 3, we can say that, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.