Mathematics in Ancient and Medieval India c6.s10.6

c6.s10.6

¹ Brahmagupta gives the general solution ( ch. 12, v. 33 ) for the rightangled triangle a² + b² = c² as a = 2pq, b = p² - q², c = p² + q². His other solution ( ch. 12, v. 35 ) for the given side a is : a, ½((a²)/p - p), ½((a²)/p + p) where p is any rational number. Mahāvīra’s solution: a, 2qa/(q² - 1), ((q² + 1)/(q² - 1))a. Putting p = ((q - 1)/(q + 1))a in Brahmagupta’s second result we get Mahāvīra’s right-angled triangle. The same solution is also given by Bhāskara II ( *Līlāvatī*, rule 139 ). Mahāvīra ( *Gaṇitasārasaṃgraha*, ch. 7, v. 222½ ) constructs another rightangled triangle for the known hypotenuse c as : c, 2pqc/(p² + q²) ((p² - q²)/(p² + q²))c, Bhāskara’s solution ( *Līlāvatī*, v. 142 ) is : c, 2nc/(n² + 1), ((n² - 1)/(n² + 1))c. This readily follows from Mahāvīra’s solution by putting p/q = n. Dickson is evidently unaware of the solution given by Brahmagupta and Mahāvīra, for he ascribes the Brahmagupta and Mahāvīra results to Leonardo Fibonacci ( c. 1202 A.D. ) and Vieta ( 1580 A.D. ) Dickson — *History of Theory of Numbers*, 2, p. 192–3 ).