Question

Find the pythagorean triplet whose smallest number is $20$

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:$2m=20\Rightarrow m=\frac{20}{2}=10$

$\Rightarrow m^2-1={10}^2-1=100-1=99$

and $m^2+1={10}^2+1=100+1=101$

$\therefore$ The triplet is $(20,99,101)$