If is non-primitive, and the integers have — greater than 1 — as a common factor, then is a Baudhayana triple? Check this statement for . Justify this statement.

Understand the Question A Baudhayana triple (Pythagorean triple) is a set of positive integers satisfying . A triple is non-primitive if the numbers share a common factor . Dividing each term by produces the set . We test this statement with and prove it generally using algebra. Step 1 · Check for the Triple Check if is a Baudhayana triple: Since , is a Baudhayana triple. The common factor of and greater than is . Dividing each term by : Now check if the resulting triple is a Baudhayana triple: Since , is a Baudhayana triple. Thus, the statement holds true for . Step 2 · Justify the General Statement Let be a non-primitive Baudhayana triple, so: Since is a common factor of and , write: where are positive integers. Substitute these into : Dividing both sides by (since ): Therefore, satisfies the condition and is a Baudhayana triple. Answer > Yes, is a Baudhayana triple. Common Mistakes Dividing by instead of : When squaring each side of the equation, the scale factor becomes . Dividing the equation by proves the relation. Common Factor Requirement: The integer must be a common factor of all three terms to ensure that are all integers.

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Question 28

If $(a, b, c)$ is non-primitive, and the integers have $f$ — greater than 1 — as a common factor, then is $\left(\dfrac{a}{f}, \dfrac{b}{f}, \dfrac{c}{f}\right)$ a Baudhayana triple? Check this statement for $(9, 12, 15)$ . Justify this statement.

Solution

Understand the Question

A Baudhayana triple (Pythagorean triple) is a set of positive integers $(a, b, c)$ satisfying $a^2 + b^2 = c^2$ .
A triple is non-primitive if the numbers share a common factor $f > 1$ .
Dividing each term by $f$ produces the set $\left(\dfrac{a}{f}, \dfrac{b}{f}, \dfrac{c}{f}\right)$ . We test this statement with $(9, 12, 15)$ and prove it generally using algebra.

Step 1 · Check for the Triple $(9, 12, 15)$

Diagram 1

Check if $(9, 12, 15)$ is a Baudhayana triple:

\begin{aligned} 9^2 + 12^2 &= 81 + 144 \\ &= 225 \end{aligned}

$15^2 = 225$

Since $9^2 + 12^2 = 15^2$ , $(9, 12, 15)$ is a Baudhayana triple.

The common factor of $9, 12,$ and $15$ greater than $1$ is $f = 3$ .

Dividing each term by $f = 3$ :

\begin{aligned} \dfrac{9}{3} &= 3 \\[0.6em] \dfrac{12}{3} &= 4 \\[0.6em] \dfrac{15}{3} &= 5 \end{aligned}

Now check if the resulting triple $(3, 4, 5)$ is a Baudhayana triple:

\begin{aligned} 3^2 + 4^2 &= 9 + 16 \\ &= 25 \end{aligned}

$5^2 = 25$

Since $3^2 + 4^2 = 5^2$ , $(3, 4, 5)$ is a Baudhayana triple. Thus, the statement holds true for $(9, 12, 15)$ .

Step 2 · Justify the General Statement

Let $(a, b, c)$ be a non-primitive Baudhayana triple, so: $a^2 + b^2 = c^2$

Since $f > 1$ is a common factor of $a, b,$ and $c$ , write: $a = f \cdot a', \quad b = f \cdot b', \quad c = f \cdot c'$ where $a', b', c'$ are positive integers.

Substitute these into $a^2 + b^2 = c^2$ : $(f \cdot a')^2 + (f \cdot b')^2 = (f \cdot c')^2$

$f^2 \cdot (a')^2 + f^2 \cdot (b')^2 = f^2 \cdot (c')^2$

$f^2 \cdot ((a')^2 + (b')^2) = f^2 \cdot (c')^2$

Dividing both sides by $f^2$ (since $f > 1 \implies f^2 \ne 0$ ): $\dfrac{f^2 \cdot ((a')^2 + (b')^2)}{f^2} = \dfrac{f^2 \cdot (c')^2}{f^2}$

$(a')^2 + (b')^2 = (c')^2$

$\left(\dfrac{a}{f}\right)^2 + \left(\dfrac{b}{f}\right)^2 = \left(\dfrac{c}{f}\right)^2$

Therefore, $\left(\dfrac{a}{f}, \dfrac{b}{f}, \dfrac{c}{f}\right)$ satisfies the condition and is a Baudhayana triple.

Answer

Yes, $\left(\dfrac{a}{f}, \dfrac{b}{f}, \dfrac{c}{f}\right)$ is a Baudhayana triple.

Common Mistakes

Dividing by $f$ instead of $f^2$ : When squaring each side of the equation, the scale factor becomes $f^2$ . Dividing the equation by $f^2$ proves the relation.
Common Factor Requirement: The integer $f$ must be a common factor of all three terms $(a, b, c)$ to ensure that $\dfrac{a}{f}, \dfrac{b}{f}, \dfrac{c}{f}$ are all integers.

Question 28

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