4. 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.

Solutions: Let x be any positive integer and y = 3. By Euclid’s division algorithm, then, 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 Now as per the question given, by squaring 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 equation 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.