Squares and Square Roots | FIO

Question 8

Find 5 more Baudhāyana triples using this idea.

Solution

Understand the Question

The sum of the first $k$ consecutive odd numbers is $k^2$ : $1 + 3 + 5 + \dots + (2k - 1) = k^2$
Adding the next consecutive odd number $(2k + 1)$ gives: $k^2 + (2k + 1) = (k + 1)^2$
If the added odd number $(2k + 1)$ is also a perfect square $m^2$ (where $m$ is an odd number $> 1$ ), it forms a Baudhāyana (Pythagorean) triple $(k, m, k + 1)$ where $k = \dfrac{m^2 - 1}{2}$ .
We can generate 5 triples by substituting consecutive odd integers $m = 7, 9, 11, 13, 15$ .

Step 1 · Find First Baudhāyana Triple (for $m = 7$ )

For $m = 7$ , set $2k + 1 = m^2$

\begin{aligned} 2k + 1 &= 7^2 \\[0.6em] 2k + 1 &= 49 \\[0.6em] 2k &= 49 - 1 \\[0.6em] 2k &= 48 \\[0.6em] k &= \dfrac{48}{2} = 24 \end{aligned}

Sum of the first $24$ odd numbers: $1 + 3 + \dots + (2 \times 24 - 1) = 1 + 3 + \dots + 47 = 24^2$

Adding $7^2$ :

\begin{aligned} 24^2 + 7^2 &= (24 + 1)^2 \\[0.6em] 24^2 + 7^2 &= 25^2 \end{aligned}

Triple: $(24, 7, 25)$

Step 2 · Find Second Baudhāyana Triple (for $m = 9$ )

For $m = 9$ , set $2k + 1 = m^2$

\begin{aligned} 2k + 1 &= 9^2 \\[0.6em] 2k + 1 &= 81 \\[0.6em] 2k &= 81 - 1 \\[0.6em] 2k &= 80 \\[0.6em] k &= \dfrac{80}{2} = 40 \end{aligned}

Sum of the first $40$ odd numbers: $1 + 3 + \dots + (2 \times 40 - 1) = 1 + 3 + \dots + 79 = 40^2$

Adding $9^2$ :

\begin{aligned} 40^2 + 9^2 &= (40 + 1)^2 \\[0.6em] 40^2 + 9^2 &= 41^2 \end{aligned}

Triple: $(40, 9, 41)$

Step 3 · Find Third Baudhāyana Triple (for $m = 11$ )

For $m = 11$ , set $2k + 1 = m^2$

\begin{aligned} 2k + 1 &= 11^2 \\[0.6em] 2k + 1 &= 121 \\[0.6em] 2k &= 121 - 1 \\[0.6em] 2k &= 120 \\[0.6em] k &= \dfrac{120}{2} = 60 \end{aligned}

Sum of the first $60$ odd numbers: $1 + 3 + \dots + (2 \times 60 - 1) = 1 + 3 + \dots + 119 = 60^2$

Adding $11^2$ :

\begin{aligned} 60^2 + 11^2 &= (60 + 1)^2 \\[0.6em] 60^2 + 11^2 &= 61^2 \end{aligned}

Triple: $(60, 11, 61)$

Step 4 · Find Fourth Baudhāyana Triple (for $m = 13$ )

For $m = 13$ , set $2k + 1 = m^2$

\begin{aligned} 2k + 1 &= 13^2 \\[0.6em] 2k + 1 &= 169 \\[0.6em] 2k &= 169 - 1 \\[0.6em] 2k &= 168 \\[0.6em] k &= \dfrac{168}{2} = 84 \end{aligned}

Sum of the first $84$ odd numbers: $1 + 3 + \dots + (2 \times 84 - 1) = 1 + 3 + \dots + 167 = 84^2$

Adding $13^2$ :

\begin{aligned} 84^2 + 13^2 &= (84 + 1)^2 \\[0.6em] 84^2 + 13^2 &= 85^2 \end{aligned}

Triple: $(84, 13, 85)$

Step 5 · Find Fifth Baudhāyana Triple (for $m = 15$ )

For $m = 15$ , set $2k + 1 = m^2$

\begin{aligned} 2k + 1 &= 15^2 \\[0.6em] 2k + 1 &= 225 \\[0.6em] 2k &= 225 - 1 \\[0.6em] 2k &= 224 \\[0.6em] k &= \dfrac{224}{2} = 112 \end{aligned}

Sum of the first $112$ odd numbers: $1 + 3 + \dots + (2 \times 112 - 1) = 1 + 3 + \dots + 223 = 112^2$

Adding $15^2$ :

\begin{aligned} 112^2 + 15^2 &= (112 + 1)^2 \\[0.6em] 112^2 + 15^2 &= 113^2 \end{aligned}

Triple: $(112, 15, 113)$

Answer

$(24, 7, 25),\ (40, 9, 41),\ (60, 11, 61),\ (84, 13, 85),\ (112, 15, 113)$

Common Mistakes

Using Even Values of $m$ : If $m$ is even, $m^2 - 1$ is odd, which would make $k = \dfrac{m^2 - 1}{2}$ a fraction instead of an integer.
Order of Terms: In a Baudhāyana triple $(a, b, c)$ , the largest number $c$ is always the hypotenuse ( $c = k + 1$ ), satisfying $a^2 + b^2 = c^2$ .

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