Question

Find the pythagorean triplet whose smallest number is $16$

Answer 1

For every natural number $m>2,(2m,m^2-1,m^2+1)$ is a pythagorean triplet and the smallest number is $2m$

Given:$2m=16\Rightarrow m=\frac{16}{2}=8$

$\Rightarrow m^2-1=8^2-1=64-1=63$

and $m^2+1=8^2+1=64+1=65$

$\therefore$ The triplet is $(16,63,65)$