The greatest number in a Pythagorean triplet is 101, then the Pythagorean triplet is

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

According to the problem

The greatest number in a Pythagorean triplet is 101

$m^2+1=101$

$m^2=100$

$m=10$

$\left(2m,m^2-1,m^2+1\right)$

$=\left(2\times 10,{10}^2-1,{10}^2+1\right)$

$=\left(20,99,101\right)$