Question

Write a pythagorean triplet whose one member is $14$.

Answer 1

Find a Pythagorean triplet whose one member is $14$:

Let the Pythagorean triplets be $2m,m^2-1,m^2+1$.

Take, $\begin{array}{rcl}2m& =& 14\end{array}$ then $\begin{array}{rcl}m& =& 7\end{array}$

Now,

$\begin{array}{rcl}{m}^2+1& =& 7^2+1\\ & =& 50\\ m^2-1& =& 7^2-1\\ & =& 48\end{array}$

Also,

$\begin{array}{rcl}{14}^2+{48}^2& =& 196+2304\\ & =& 2500\\ & =& {50}^2\end{array}$

Hence, the Pythagorean triplet is $14,48,50$ whose one member is $14$.