Baudhāyana triples
A right-angled triangle has one angle equal to \(90^\circ\). The side opposite the right angle is called the hypotenuse. According to the Baudhāyana-Pythagoras Theorem, if the two shorter sides have lengths \(a\) and \(b\), and the hypotenuse has length \(c\), then \(c^2 = a^2 + b^2\).
A triple of positive integers \((a, b, c)\) satisfying \(c^2 = a^2 + b^2\) is called a Baudhāyana triple. They are also called as Baudhāyana-Pythagoras triples, Pythagorean triples and Right-angled triangle triples.
In the Śulba-Sūtra (Verse 1.13), Baudhāyana lists \((3, 4, 5)\), \((5, 12, 13)\), \((8, 15, 17)\), \((7, 24, 25)\), \((12, 35, 37)\) and \((15, 36, 39)\).
Verifying Baudhāyana triple
Consider the Pythagorean triplet \((3, 4, 5)\).
Let \(a = 3\), \(b = 4\) and \(c = 5\).
Applying the Baudhāyana - Pythagoras theorem, we have:
\(a^2 + b^2 = c^2\)
LHS: \(a^2 + b^2 = 3^2 + 4^2\) \(= 9 + 16 = 25\)
RHS: \(c^2 = 5^2 = 25\)
Since LHS \(=\) RHS, \((3, 4, 5)\) is a Baudhāyana triple.
Triples with all numbers less than or equal to \(20\)
| Triple | Multiplied by |
| \((3, 4, 5)\) | \(1\) |
| \((6, 8, 10)\) | \(2 \times (3, 4, 5)\) |
| \((9, 12, 15)\) | \(3 \times (3, 4, 5)\) |
| \((12, 16, 20)\) | \(4 \times (3, 4, 5)\) |
Pattern: These are multiples of \((3, 4, 5)\). Also, \((30, 40, 50)\) and \((300, 400, 500)\) are triples, since they are \(10 \times (3, 4, 5)\) and \(100 × (3, 4, 5)\).
Scaled version of the triple
Conjecture: \((3k, 4k, 5k)\) is a triple for every positive integer \(k\).
Proof:
Consider the triple \((3k, 4k, 5k)\).
Let \(a = 3k\), \(b = 4k\) and \(c = 5k\).
Applying the Baudhāyana - Pythagoras theorem, we have:
\(a^2 + b^2 = c^2\)
LHS: \(a^2 + b^2 = (3k)^2 + (4k)^2\) \(= 9k^2 + 16k^2 = 25k^2\)
RHS: \(c^2 = (5k)^2 = 25k^2\)
Since LHS \(=\) RHS, \((3k, 4k, 5k)\) is a Baudhāyana triple.
Statement: If \((a, b, c)\) is a Baudhāyana triple, then \((ka, kb, kc)\) is also a Baudhāyana triple where \(k\) is any positive integer.
To prove: \((ka)^2 + (kb)^2 = (kc)^2\)
Proof:
Consider the triple \((ka, kb, kc)\).
\((ka)^2 + (kb)^2 = k^2a^2 + k^2b^2\)
\(= k^2(a^2 + b^2)\)
\(= k^2c^2\) [Since \(a^2 + b^2 = c^2\)]
\(= (kc)^2\)
Thus, \((ka, kb, kc)\) is a Baudhāyana triple if \((a, b, c)\) is a Baudhayana triple. It is also called the scaled version of \((a, b, c)\).
Primitive triples
A Baudhāyana triple with no common factor greater than \(1\) is primitive.
Example:
\((3, 4, 5)\) and \((5, 12, 13)\) are primitive.
\((12, 16, 20)\) is not, because it has the common factor \(4\).
Reverse scaling: If \((a, b, c)\) has a common factor \(f > 1\), then \((\frac{a}{f}, \frac{b}{f}, \frac{c}{f})\) is also a triple.
Proof:
\((\frac{a}{f})² + (\frac{b}{f})² = \frac{a^2}{f^2} + \frac{b^2}{f^2}\)
\(= \frac{(a^2 + b^2)}{f^2}\)
\(= \frac{c^2}{f^2}\)
\(= \frac{c}{f}\)
Thus, \((\frac{a}{f}, \frac{b}{f}, \frac{c}{f})\) is also a triple.
Example:
\((16, 30, 34)\) \(\div 2\) \(= (8, 15, 17)\)
Generating primitive triples from odd numbers
Step 1: the odd-number fact.
- \(1 = 1²\), \(1 + 3 = 2²\), \(1 + 3 + 5 = 3²\)
- The sum of the first \(n\) odd numbers is \(n²\).
- The \(n\)th odd number is \(2n − 1\).
Step 2: the key identity.
- \(1 + 3 + … + (2n − 3)\) is the sum of the first \((n − 1)\) odd numbers, which is \((n − 1)²\).
- Adding the \(n\)th odd number gives: \((n − 1)² + (2n − 1) = n²\)
- This can also be checked by expanding \((n − 1)² = n² − 2n + 1\) and adding \(2n − 1\).
Step 3: make \(2n − 1\) a perfect square. If \(2n − 1 = m²\) (\(m\) is an odd number), we get a sum of two squares equal to a square.
Examples:
- \(9 = 2 × 5 − 1\), so \(n = 5\): \(4² + 3² = 5²\) \(→ (3, 4, 5)\)
- \(49 = 2 × 25 − 1\), so \(n = 25\): \(24² + 7² = 25²\) \(→ (7, 24, 25)\)