Question 20

A Baudhāyana triple is a set of three positive whole numbers $a, b, c$. They satisfy the equation $a^2 + b^2 = c^2$.

Step 1 — Understanding Baudhāyana Triples

We are looking for three positive whole numbers. Let these numbers be $a, b,$ and $c$. They must satisfy the relationship $a^2 + b^2 = c^2$. This is the same condition as for Pythagorean triples. All numbers in the triple must be less than or equal to 20.

Step 2 — Generating Primitive Triples

We can use a formula to generate these triples. Let $m$ and $n$ be two positive integers. We need $m > n > 0$. Also, $m$ and $n$ must not have any common factors. This means they are coprime. One of $m$ or $n$ must be an even number. The primitive triples $(a, b, c)$ are given by these formulas:

$a = m^2 - n^2$

$b = 2mn$

$c = m^2 + n^2$

We will find triples where $c \le 20$. So, $m^2 + n^2$ must be less than or equal to 20.

Let us test values for $m$ and $n$:

• Case 1: $m=2, n=1$ (Here $m>n$, $m, n$ are coprime, and $m$ is even) $a = 2^2 - 1^2 = 4 - 1 = \mathbf{3}$ $b = 2 \times 2 \times 1 = \mathbf{4}$ $c = 2^2 + 1^2 = 4 + 1 = \mathbf{5}$ The triple is (3, 4, 5). All numbers are $\le 20$.

• Case 2: $m=3, n=2$ (Here $m>n$, $m, n$ are coprime, and $n$ is even) $a = 3^2 - 2^2 = 9 - 4 = \mathbf{5}$ $b = 2 \times 3 \times 2 = \mathbf{12}$ $c = 3^2 + 2^2 = 9 + 4 = \mathbf{13}$ The triple is (5, 12, 13). All numbers are $\le 20$.

• Case 3: $m=4, n=1$ (Here $m>n$, $m, n$ are coprime, and $m$ is even) $a = 4^2 - 1^2 = 16 - 1 = \mathbf{15}$ $b = 2 \times 4 \times 1 = \mathbf{8}$ $c = 4^2 + 1^2 = 16 + 1 = \mathbf{17}$ The triple is (8, 15, 17) (we list $a \le b$). All numbers are $\le 20$.

• Case 4: $m=4, n=3$ (Here $m>n$, $m, n$ are coprime, and $m$ is even) $a = 4^2 - 3^2 = 16 - 9 = \mathbf{7}$ $b = 2 \times 4 \times 3 = \mathbf{24}$ $c = 4^2 + 3^2 = 16 + 9 = \mathbf{25}$ Here, $b=24$ and $c=25$ are greater than 20. So this triple is not included.

• Case 5: $m=5, n=2$ (Here $m>n$, $m, n$ are coprime, and $n$ is even) $a = 5^2 - 2^2 = 25 - 4 = \mathbf{21}$ Here, $a=21$ is greater than 20. So this triple is not included.

We can stop here. If $m$ increases further, $c = m^2 + n^2$ will also increase. For example, if $m=5, n=1$, $c = 5^2+1^2 = 26$, which is greater than 20.

The primitive Baudhāyana triples with $c \le 20$ are:

$\boxed{\text{(3, 4, 5)}}$ $\boxed{\text{(5, 12, 13)}}$ $\boxed{\text{(8, 15, 17)}}$

Step 3 — Generating Non-Primitive Triples

Non-primitive triples are multiples of primitive triples. Let $(a, b, c)$ be a primitive triple. Then $(ka, kb, kc)$ is a non-primitive triple for any integer $k > 1$. All numbers in the triple must be less than or equal to 20.

Let us check multiples of the primitive triples we found:

• For (3, 4, 5):

  • Multiply by $k=2$: $(2 \times 3, 2 \times 4, 2 \times 5) = \mathbf{(6, 8, 10)}$. All numbers are $\le 20$.
  • Multiply by $k=3$: $(3 \times 3, 3 \times 4, 3 \times 5) = \mathbf{(9, 12, 15)}$. All numbers are $\le 20$.
  • Multiply by $k=4$: $(4 \times 3, 4 \times 4, 4 \times 5) = \mathbf{(12, 16, 20)}$. All numbers are $\le 20$.
  • Multiply by $k=5$: $(5 \times 3, 5 \times 4, 5 \times 5) = (15, 20, 25)$. Here, $c=25$ is greater than 20. So we stop.

• For (5, 12, 13):

  • Multiply by $k=2$: $(2 \times 5, 2 \times 12, 2 \times 13) = (10, 24, 26)$. Here, $b=24$ and $c=26$ are greater than 20. So we stop.

• For (8, 15, 17):

  • Multiply by $k=2$: $(2 \times 8, 2 \times 15, 2 \times 17) = (16, 30, 34)$. Here, $b=30$ and $c=34$ are greater than 20. So we stop.

Answer

The Baudhāyana triples with numbers less than or equal to 20 are:

(i) (3, 4, 5) (ii) (5, 12, 13) (iii) (6, 8, 10) (iv) (8, 15, 17) (v) (9, 12, 15) (vi) (12, 16, 20)