Question 8

We can find Baudhāyana triples by using the property that the sum of the first 'n' odd numbers is $n^2$.

Step 1 — Understanding the Pattern

We know that the sum of the first $n$ odd numbers is $n^2$. For example, $1+3 = 4 = 2^2$. Also, $1+3+5 = 9 = 3^2$. So, we can write this as $1 + 3 + 5 + ... + (2n-1) = n^2$. A Baudhāyana triple $(a, b, c)$ satisfies the equation $a^2 + b^2 = c^2$. Let us use the sum of odd numbers to form one part of this equation. We can take $a^2$ to be the sum of the first $k$ odd numbers. So, $a^2 = 1 + 3 + 5 + ... + (2k-1)$. This means $a^2 = k^2$. Now, we add the next odd number to this sum. The next odd number after $(2k-1)$ is $(2k-1) + 2 = 2k+1$. So, we have the expression $k^2 + (2k+1)$. We know from algebra that $k^2 + 2k + 1$ is equal to $(k+1)^2$. So, we have the equation $k^2 + (2k+1) = (k+1)^2$. For this to be a Baudhāyana triple, the term $(2k+1)$ must be a perfect square. Let $2k+1 = m^2$ for some integer $m$. Then our Baudhāyana triple will be $(k, m, k+1)$. For $k$ to be a whole number, $m^2-1$ must be an even number. This means $m^2$ must be an odd number. So, $m$ must be an odd number. We need $m$ to be greater than 1 for $k$ to be a positive integer. The smallest odd number for $m$ is 3, which gives the triple $(4, 3, 5)$. The problem asks for 5 Baudhāyana triples. We will find them using consecutive odd values for $m$, starting from $m=7$.

Step 2 — First Baudhāyana Triple (for $m=7$)

We choose the odd number 7 for $m$. We set $2k+1$ equal to $m^2$.

$2k+1 = 7^2$

$2k+1 = 49$

$2k = 49 - 1$

$2k = 48$

$k = \frac{48}{2}$

$\boxed{k = 24}$

Now we write the sum of odd numbers. The sum of the first $k$ odd numbers is $k^2$. So, $1 + 3 + ... + (2k-1)$ is $1 + 3 + ... + (2 \times 24 - 1)$.

$1 + 3 + ... + 47 = 24^2$

We add $m^2$ to $k^2$ to get $(k+1)^2$.

$24^2 + 7^2 = (24+1)^2$

$24^2 + 7^2 = 25^2$

This gives us our first Baudhāyana triple.

$\boxed{\text{(24, 7, 25)}}$

Step 3 — Second Baudhāyana Triple (for $m=9$)

We choose the next odd number 9 for $m$. We set $2k+1$ equal to $m^2$.

$2k+1 = 9^2$

$2k+1 = 81$

$2k = 81 - 1$

$2k = 80$

$k = \frac{80}{2}$

$\boxed{k = 40}$

Now we write the sum of odd numbers. The sum of the first $k$ odd numbers is $k^2$. So, $1 + 3 + ... + (2k-1)$ is $1 + 3 + ... + (2 \times 40 - 1)$.

$1 + 3 + ... + 79 = 40^2$

We add $m^2$ to $k^2$ to get $(k+1)^2$.

$40^2 + 9^2 = (40+1)^2$

$40^2 + 9^2 = 41^2$

This gives us our second Baudhāyana triple.

$\boxed{\text{(40, 9, 41)}}$

Step 4 — Third Baudhāyana Triple (for $m=11$)

We choose the next odd number 11 for $m$. We set $2k+1$ equal to $m^2$.

$2k+1 = 11^2$

$2k+1 = 121$

$2k = 121 - 1$

$2k = 120$

$k = \frac{120}{2}$

$\boxed{k = 60}$

Now we write the sum of odd numbers. The sum of the first $k$ odd numbers is $k^2$. So, $1 + 3 + ... + (2k-1)$ is $1 + 3 + ... + (2 \times 60 - 1)$.

$1 + 3 + ... + 119 = 60^2$

We add $m^2$ to $k^2$ to get $(k+1)^2$.

$60^2 + 11^2 = (60+1)^2$

$60^2 + 11^2 = 61^2$

This gives us our third Baudhāyana triple.

$\boxed{\text{(60, 11, 61)}}$

Step 5 — Fourth Baudhāyana Triple (for $m=13$)

We choose the next odd number 13 for $m$. We set $2k+1$ equal to $m^2$.

$2k+1 = 13^2$

$2k+1 = 169$

$2k = 169 - 1$

$2k = 168$

$k = \frac{168}{2}$

$\boxed{k = 84}$

Now we write the sum of odd numbers. The sum of the first $k$ odd numbers is $k^2$. So, $1 + 3 + ... + (2k-1)$ is $1 + 3 + ... + (2 \times 84 - 1)$.

$1 + 3 + ... + 167 = 84^2$

We add $m^2$ to $k^2$ to get $(k+1)^2$.

$84^2 + 13^2 = (84+1)^2$

$84^2 + 13^2 = 85^2$

This gives us our fourth Baudhāyana triple.

$\boxed{\text{(84, 13, 85)}}$

Step 6 — Fifth Baudhāyana Triple (for $m=15$)

We choose the next odd number 15 for $m$. We set $2k+1$ equal to $m^2$.

$2k+1 = 15^2$

$2k+1 = 225$

$2k = 225 - 1$

$2k = 224$

$k = \frac{224}{2}$

$\boxed{k = 112}$

Now we write the sum of odd numbers. The sum of the first $k$ odd numbers is $k^2$. So, $1 + 3 + ... + (2k-1)$ is $1 + 3 + ... + (2 \times 112 - 1)$.

$1 + 3 + ... + 223 = 112^2$

We add $m^2$ to $k^2$ to get $(k+1)^2$.

$112^2 + 15^2 = (112+1)^2$

$112^2 + 15^2 = 113^2$

This gives us our fifth Baudhāyana triple.

$\boxed{\text{(112, 15, 113)}}$

Answer

(i) (24, 7, 25) (ii) (40, 9, 41) (iii) (60, 11, 61) (iv) (84, 13, 85) (v) (112, 15, 113)