Question 9

We will first understand what Baudhāyana triples and primitive triples are. Then, we will find the general form of triples generated by the given method. Finally, we will check the greatest common factor (HCF) of these triples to see if they are primitive or non-primitive.

Step 1 — Understanding Baudhāyana Triples

A Baudhāyana triple is a set of three positive whole numbers. Let these numbers be $a$, $b$, and $c$. They satisfy the rule $a^2 + b^2 = c^2$. These triples describe the side lengths of a right-angled triangle. The number $c$ is always the longest side, called the hypotenuse.

Step 2 — Understanding Primitive Triples

A Baudhāyana triple $(a, b, c)$ is called primitive. This means the greatest common factor (HCF) of $a$, $b$, and $c$ is 1. If the HCF of $a$, $b$, and $c$ is greater than 1, the triple is non-primitive. For example, $(6, 8, 10)$ is a non-primitive triple. The HCF of 6, 8, and 10 is 2.

Step 3 — Identifying the Method for Generating Triples

The hint tells us that one of the smaller side lengths is one less than the hypotenuse. Let the three sides of the triple be $a$, $b$, and $c$. Let $c$ be the hypotenuse. So, one of the smaller sides, say $b$, is $c-1$. We know that $a^2 + b^2 = c^2$. Let us substitute $b = c-1$ into this equation. $a^2 + (c-1)^2 = c^2$ We expand $(c-1)^2$ using the identity $(x-y)^2 = x^2 - 2xy + y^2$. $a^2 + c^2 - 2c + 1 = c^2$ Now, we subtract $c^2$ from both sides of the equation. $a^2 - 2c + 1 = 0$ We want to find $c$, so we rearrange the equation. $2c = a^2 + 1$ $c = \frac{a^2 + 1}{2}$ Now we can find the value of $b$ using $b = c-1$. $b = \frac{a^2 + 1}{2} - 1$ $b = \frac{a^2 + 1 - 2}{2}$ $b = \frac{a^2 - 1}{2}$ For $b$ and $c$ to be whole numbers, $a^2+1$ and $a^2-1$ must both be even. This means $a^2$ must be an odd number. If $a^2$ is odd, then $a$ must also be an odd whole number. So, the method generates Baudhāyana triples $(a, \frac{a^2-1}{2}, \frac{a^2+1}{2})$ for any odd whole number $a$.

Step 4 — Checking the HCF of Generated Triples

Let the generated triple be $(a, b, c)$. Here, $a$ is an odd whole number. The second side is $b = \frac{a^2-1}{2}$. The third side (hypotenuse) is $c = \frac{a^2+1}{2}$. We need to find the greatest common factor (HCF) of $a$, $b$, and $c$. Let us find the difference between $c$ and $b$. $c - b = \frac{a^2+1}{2} - \frac{a^2-1}{2}$ $= \frac{(a^2+1) - (a^2-1)}{2}$ $= \frac{a^2+1 - a^2+1}{2}$ $= \frac{2}{2}$

$\boxed{= 1}$ So, we have a triple $(a, b, c)$ where $c-b=1$. The HCF of $(a, b, c)$ is the same as the HCF of $(a, b, c-b)$. Therefore, HCF$(a, b, c) = \text{HCF}(a, b, 1)$. The greatest common factor of any set of numbers that includes 1 is always 1. So, the HCF of $a$, $b$, and $c$ is always 1.

Step 5 — Generating Example Triples

Let us use an odd number, $a = 7$. The first side is $a = \mathbf{7}$. The second side is $b = \frac{7^2-1}{2}$. $b = \frac{49-1}{2}$ $= \frac{48}{2}$

$\boxed{= 24}$ The third side (hypotenuse) is $c = \frac{7^2+1}{2}$. $c = \frac{49+1}{2}$ $= \frac{50}{2}$ $\boxed{= 25}$ The Baudhāyana triple is (7, 24, 25). The HCF of 7, 24, and 25 is 1. This is a primitive triple.

Let us use another odd number, $a = 9$. The first side is $a = \mathbf{9}$. The second side is $b = \frac{9^2-1}{2}$. $b = \frac{81-1}{2}$ $= \frac{80}{2}$

$\boxed{= 40}$ The third side (hypotenuse) is $c = \frac{9^2+1}{2}$. $c = \frac{81+1}{2}$ $= \frac{82}{2}$ $\boxed{= 41}$ The Baudhāyana triple is (9, 40, 41). The HCF of 9, 40, and 41 is 1. This is also a primitive triple.

Answer

The method yields primitive Baudhāyana triples.

No, this method does not yield non-primitive Baudhāyana triples.