Q.1: Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m. 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 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 + 1 2 + 2 × 3q × 1 = 9q 2 + 1 + 6q = 3(3q 2 + 2q) + 1 Substituting 3q 2 +2q = m we get, x 2 = 3m + 1 ……………………………. (2) x 2 = (3q + 2) 2 = (3q) 2 + 2 2 + 2 × 3q × 2 = 9q 2 + 4 + 12q = 3(3q 2 + 4q + 1) + 1 Again, substituting 3q 2 + 4q + 1 = m, we 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...
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