Find 5 more Baudhāyana triples using this idea.
Five more Baudhāyana triples using the odd square method:
Method: When the nth odd number (2n-1) is a perfect square, it generates a Baudhāyana triple where one smaller side is one less than the hypotenuse.
Step 1: Find odd square numbers: 49, 81, 121, 169, 225
Step 2: For 49 = 7² (the 25th odd number): (25-1)² + 7² = 25² ∴ (7, 24, 25)
Step 3: For 81 = 9² (the 41st odd number): (41-1)² + 9² = 41² ∴ (9, 40, 41)
Step 4: For 121 = 11² (the 61st odd number): (61-1)² + 11² = 61² ∴ (11, 60, 61)
Step 5: For 169 = 13² (the 85th odd number): (85-1)² + 13² = 85² ∴ (13, 84, 85)
Step 6: For 225 = 15² (the 113th odd number): (113-1)² + 15² = 113² ∴ (15, 112, 113)
Answer: Five more Baudhāyana triples: (7, 24, 25), (9, 40, 41), (11, 60, 61), (13, 84, 85), (15, 112, 113)
Explanation
The chapter explains that when the nth odd number (2n-1) is a perfect square, we get a Baudhāyana triple. For example, 49 = 7² is the 25th odd number, giving 24² + 7² = 25². Similarly, 81 = 9² is the 41st odd number (41 = (81+1)/2), giving 40² + 9² = 41². The pattern continues: 121 = 11² gives (11, 60, 61), 169 = 13² gives (13, 84, 85), and 225 = 15² gives (15, 112, 113).
Solution Steps
Step 1: Find odd square numbers: 49, 81, 121, 169, 225
Step 2: For 49 = 7², it's the 25th odd number: (25-1)² + 7² = 25² → (7, 24, 25)
Step 3: For 81 = 9², it's the 41st odd number: (41-1)² + 9² = 41² → (9, 40, 41)
Step 4: For 121 = 11², it's the 61st odd number: (61-1)² + 11² = 61² → (11, 60, 61)
Step 5: For 169 = 13², it's the 85th odd number: (85-1)² + 13² = 85² → (13, 84, 85)
Step 6: For 225 = 15², it's the 113th odd number: (113-1)² + 15² = 113² → (15, 112, 113)