Mathematics in Ancient and Medieval India c5.s13.21

c5.s13.21

Now 1001 = 31² + 40 and 2 × 31 + 1–40 = 23, not a perfect square. ∴ 23 + (2.31 + 3) + (2.31 + 5) + …… = k², say. i.e. 23 + 65 + 67 + …… + 89 = k² = 32². hence by the rule, 1001 + k² = 1001 + 32² = 45² or, 1001 = 45²–32² = 77 × 13 = 7 × 11 × 13 The possible divisors are : 7, 11, 13, 77, 91, 143, 1001