Question

Write a Pythagorean triplet whose smallest number is $8$ and verify your answer.

Answer 1

Find a Pythagorean triplet whose smallest number is $8$:
Let the Pythagorean triplets be $2m,m^2-1,m^2+1$.

Take $\begin{array}{rcl}2m& =& 8\end{array}$ then $\begin{array}{rcl}m& =& 4\end{array}$

Now,

$\begin{array}{rcl}{m}^2-1& =& 4^2-1\\ & =& 15\\ m^2+1& =& 4^2+1\\ & =& 17\end{array}$

Pythagorean triplet is $8,15,17$.

This can be verified as follows:

$\begin{array}{rcl}{8}^2+{15}^2& =& 64+225\\ & =& 289\\ & =& {17}^2\\ \therefore 8^2+{15}^2& =& {17}^2\end{array}$

Hence, the Pythagorean triplet is $8,15,17$.