Question 26

A Baudhayana triple is a set of three positive whole numbers $(a, b, c)$ such that $a^2 + b^2 = c^2$. A primitive Baudhayana triple is one where $a, b, c$ have no common factors other than 1.

Step 1 — Verify if $(5, 12, 13)$ is a Baudhayana triple

Let us check if the numbers $(5, 12, 13)$ satisfy the condition $a^2 + b^2 = c^2$. Here, we have $a = 5$, $b = 12$, and $c = 13$. First, we will calculate the sum of the squares of the first two numbers. $a^2 + b^2 = 5^2 + 12^2$ $= 25 + 144$ $= 169$ Next, we will calculate the square of the third number. $c^2 = 13^2$ $= 169$ Since $a^2 + b^2 = c^2$, the numbers $(5, 12, 13)$ form a Baudhayana triple.

$\boxed{\text{Yes, (5, 12, 13) is a Baudhayana triple.}}$

Step 2 — Check for primitivity of $(5, 12, 13)$

A Baudhayana triple is called primitive if its three numbers have no common factors other than 1. Let us find the factors of each number in the triple $(5, 12, 13)$. The factors of 5 are 1, 5. The factors of 12 are 1, 2, 3, 4, 6, 12. The factors of 13 are 1, 13. The only common factor for 5, 12, and 13 is 1. So, the triple $(5, 12, 13)$ is a primitive Baudhayana triple.

$\boxed{\text{Yes, (5, 12, 13) is a primitive Baudhayana triple.}}$

Step 3 — Find other primitive Baudhayana triples with numbers less than or equal to 20

We can find primitive Baudhayana triples using a special formula. Let $m$ and $n$ be two positive whole numbers. These numbers must satisfy three conditions:

1. $m$ must be greater than $n$ ($m > n$).
2. $m$ and $n$ must have no common factors other than 1 (they are coprime).
3. One of $m$ or $n$ must be an even number. If these conditions are met, the primitive Baudhayana triple $(a, b, c)$ can be found using these formulas: $a = m^2 - n^2$ $b = 2mn$ $c = m^2 + n^2$ We need to find triples where all three numbers ($a, b, c$) are less than or equal to 20.

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

Case 1: Let $m = 2$, $n = 1$. These values satisfy all conditions: $2 > 1$, 2 and 1 are coprime, and 2 is an even number. $a = m^2 - n^2 = 2^2 - 1^2 = 4 - 1 = \mathbf{3}$ $b = 2mn = 2 \times 2 \times 1 = \mathbf{4}$ $c = m^2 + n^2 = 2^2 + 1^2 = 4 + 1 = \mathbf{5}$ The triple is $(3, 4, 5)$. All numbers (3, 4, 5) are less than or equal to 20. This is a primitive Baudhayana triple.

Case 2: Let $m = 3$, $n = 2$. These values satisfy all conditions: $3 > 2$, 3 and 2 are coprime, and 2 is an even number. $a = m^2 - n^2 = 3^2 - 2^2 = 9 - 4 = \mathbf{5}$ $b = 2mn = 2 \times 3 \times 2 = \mathbf{12}$ $c = m^2 + n^2 = 3^2 + 2^2 = 9 + 4 = \mathbf{13}$ The triple is $(5, 12, 13)$. All numbers (5, 12, 13) are less than or equal to 20. This is the triple given in the question.

Case 3: Let $m = 4$, $n = 1$. These values satisfy all conditions: $4 > 1$, 4 and 1 are coprime, and 4 is an even number. $a = m^2 - n^2 = 4^2 - 1^2 = 16 - 1 = \mathbf{15}$ $b = 2mn = 2 \times 4 \times 1 = \mathbf{8}$ $c = m^2 + n^2 = 4^2 + 1^2 = 16 + 1 = \mathbf{17}$ The triple is $(15, 8, 17)$. All numbers (15, 8, 17) are less than or equal to 20. This is a primitive Baudhayana triple.

Case 4: Let $m = 4$, $n = 3$. These values satisfy conditions: $4 > 3$, 4 and 3 are coprime, and 4 is an even number. $a = m^2 - n^2 = 4^2 - 3^2 = 16 - 9 = \mathbf{7}$ $b = 2mn = 2 \times 4 \times 3 = \mathbf{24}$ $c = m^2 + n^2 = 4^2 + 3^2 = 16 + 9 = \mathbf{25}$ Here, $b = 24$ and $c = 25$ are greater than 20. So, this triple is not included. If we try larger values for $m$, the value of $c = m^2 + n^2$ will become even larger than 20. For example, if $m=5$, then $m^2 = 25$, which is already greater than 20. So, we can stop here.

The primitive Baudhayana triples with numbers less than or equal to 20 are $(3, 4, 5)$, $(5, 12, 13)$, and $(8, 15, 17)$. The question asks for the "other" primitive Baudhayana triples, so we exclude $(5, 12, 13)$.

$\boxed{\text{The other primitive Baudhayana triples are (3, 4, 5) and (8, 15, 17).}}$

Answer

(i) Yes, $(5, 12, 13)$ is a primitive Baudhayana triple. (ii) The other primitive Baudhayana triples with numbers less than or equal to 20 are $(3, 4, 5)$ and $(8, 15, 17)$.