Question 28

We will check if dividing a non-primitive Baudhayana triple by its common factor still results in a Baudhayana triple.

Step 1 — Understanding Baudhayana Triples

A Baudhayana triple is a set of three positive integers $(a, b, c)$. These integers satisfy the equation $a^2 + b^2 = c^2$. For example, $(3, 4, 5)$ is a Baudhayana triple because $3^2 + 4^2 = 9 + 16 = 25 = 5^2$. A Baudhayana triple $(a, b, c)$ is non-primitive if $a, b, c$ have a common factor greater than 1. A common factor is a number that divides all three integers exactly. Let $f$ be this common factor.

Step 2 — Checking the example $(9, 12, 15)$

First, let us check if $(9, 12, 15)$ is a Baudhayana triple. We need to see if $9^2 + 12^2 = 15^2$.

$9^2 + 12^2$

$= 81 + 144$

$= 225$

$15^2$

$= 225$

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

Next, we find the common factor $f$ for $9, 12,$ and $15$. Factors of $9$ are $1, 3, 9$. Factors of $12$ are $1, 2, 3, 4, 6, 12$. Factors of $15$ are $1, 3, 5, 15$. The common factors are $1$ and $3$. The common factor $f$ greater than $1$ is $3$.

Now, we divide each number in the triple by $f = 3$. The new triple is $\left(\frac{9}{3}, \frac{12}{3}, \frac{15}{3}\right)$.

$\frac{9}{3} = 3$

$\frac{12}{3} = 4$

$\frac{15}{3} = 5$

The new triple is $(3, 4, 5)$. Let us check if $(3, 4, 5)$ is a Baudhayana triple. We need to see if $3^2 + 4^2 = 5^2$.

$3^2 + 4^2$

$= 9 + 16$

$= 25$

$5^2$

$= 25$

So, $3^2 + 4^2 = 5^2$. This means $(3, 4, 5)$ is a Baudhayana triple. For the example $(9, 12, 15)$, the statement holds true.

Step 3 — Justifying the general statement

Let $(a, b, c)$ be a non-primitive Baudhayana triple. This means $a^2 + b^2 = c^2$. Since it is non-primitive, there is a common factor $f$. We know $f$ is an integer and $f > 1$. This means $a, b,$ and $c$ are all multiples of $f$. So, we can write $a = f \cdot a'$, $b = f \cdot b'$, and $c = f \cdot c'$. Here, $a', b',$ and $c'$ are also integers. We want to check if $\left(\frac{a}{f}, \frac{b}{f}, \frac{c}{f}\right)$, which is $(a', b', c')$, is a Baudhayana triple. This means we need to check if $(a')^2 + (b')^2 = (c')^2$.

Let us substitute $a = f \cdot a'$, $b = f \cdot b'$, and $c = f \cdot c'$ into the original equation $a^2 + b^2 = c^2$.

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

We can expand the terms.

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

Now, we can take $f^2$ as a common factor on the left side.

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

Since $f > 1$, $f^2$ is not zero. So, we can divide both sides of the equation by $f^2$.

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

This simplifies to:

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

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

This shows that the new triple $(a', b', c')$ also satisfies the condition for a Baudhayana triple. So, if $(a, b, c)$ is a non-primitive Baudhayana triple, then $\left(\frac{a}{f}, \frac{b}{f}, \frac{c}{f}\right)$ is also a Baudhayana triple.

Answer

(i) Yes, if $(a, b, c)$ is non-primitive, and $f$ is a common factor greater than 1, then $\left(\frac{a}{f}, \frac{b}{f}, \frac{c}{f}\right)$ is a Baudhayana triple. (ii) For $(9, 12, 15)$, the common factor $f$ is $3$. Dividing by $3$ gives $(3, 4, 5)$, which is a Baudhayana triple ($3^2 + 4^2 = 5^2$). (iii) The statement is justified because if $a^2 + b^2 = c^2$ and $a=fa', b=fb', c=fc'$, then $(fa')^2 + (fb')^2 = (fc')^2$, which simplifies to $f^2(a'^2 + b'^2) = f^2c'^2$. Dividing by $f^2$ gives $a'^2 + b'^2 = c'^2$.