Question

Write a Pythagorean triplet whose one member is. i) $6$ ii) $14$ iii) $16$ iv) $18$

Answer 1

Step 1. Write a Pythagorean triplet whose one member is $6$ :

The Pythagorean triplet is formed by the numbers $2m,m^2-1,\mathrm{and}{\mathrm{m}}^2+1$ such that ${\left(2m\right)}^2+{\left(m^2-1\right)}^2={\left(m^2+1\right)}^2$ for any natural number$m>1$.

$\begin{array}{rcl}\mathrm{take}2m& =& 6\\ & \Rightarrow & m=3\end{array}$

Put $\begin{array}{rcl}m& =& 3\end{array}$, to calculate the other two numbers as follows:

$\begin{array}{rcl}{m}^2+1& =& 3^2+1\\ & \Rightarrow & m^2+1=10\end{array}$

And$\begin{array}{rcl}{m}^2+1& =& 3^2+1\\ & \Rightarrow & m^2+1=10\end{array}$

Hence, the Pythagorean triplet whose one member is $6$.is $\left(6,8,10\right)$.

Step 2. Write a Pythagorean triplet whose one member is$14$:

$\begin{array}{rcl}\mathrm{take}2m& =& 14\\ & \Rightarrow & m=7\end{array}$

Put $\begin{array}{rcl}m& =& 7\end{array}$, to calculate the other two numbers as follows:

$\begin{array}{rcl}{m}^2+1& =& 7^2+1\\ & \Rightarrow & m^2+1=50\end{array}$

And,

$\begin{array}{rcl}{m}^2-1& =& 7^2-1\\ & \Rightarrow & m^2-1=48\end{array}$

Hence, the Pythagorean triplet one member is $14$ is $\left(14,48,50\right)$.

Step 3. Write a Pythagorean triplet whose one member is$16$ :

$\begin{array}{rcl}\mathrm{take}2m& =& 16\\ & \Rightarrow & m=8\end{array}$

Put $\begin{array}{rcl}m& =& 8\end{array}$, to calculate the other two numbers as follows:

$\begin{array}{rcl}{m}^2+1& =& 8^2+1\\ & \Rightarrow & m^2+1=65\end{array}$

And

$\begin{array}{rcl}{m}^2+1& =& 8^2+1\\ & \Rightarrow & m^2+1=65\end{array}$

Hence, the Pythagorean triplet one member is $16$ $\left(16,63,65\right)$.

Step 4. Write a Pythagorean triplet whose one member is $18$:

$\begin{array}{rcl}\mathrm{take}2m& =& 18\\ & \Rightarrow & m=9\end{array}$

Put $\begin{array}{rcl}m& =& 9\end{array}$, to calculate the other two numbers as follows:

$\begin{array}{rcl}{m}^2-1& =& 9^2-1\\ & \Rightarrow & m^2-1=80\end{array}$

And

$\begin{array}{rcl}{m}^2+1& =& 9^2+1\\ & \Rightarrow & m^2+1=82\end{array}$

Hence, the Pythagorean triplet whose one member is $18$ $\left(18,80,82\right)$.

Hence the Pythagorean triplet including

(i) $6$is $\left(6,8,10\right)$

(ii)$14$is $\left(14,48,50\right)$

(iii)$16$ is $\left(16,63,65\right)$

(iv) $18$ is$\left(18,80,82\right)$