If one number of a Pythagorean triplet is 2m, then the other two numbers are 

A: $m,m^2+1,2m$ then smallest number is m.

Greatest number is $m^2+1$

Pythagorean triplet : ${\left(m^2+1\right)}^2={\left(m^2\right)}^2+2.m^{2.}.1+1^2=m^4+2m^2+1$

                                   $m^2+{\left(2m\right)}^2=m^2+4m^2=5m^2$

                                      $\therefore {\left(m^2+1\right)}^2\ne m^2+{\left(2m\right)}^2$

B: $m^2+1,m^2-1,2m$ then the smallest number is 2m

Greatest number is $m^2+1$

Pythagorean triplet : 

$\begin{array}{c}{\left(m^2+1\right)}^2=m^4+2m^2+1\\ {\left(m^2-1\right)}^2+{\left(2m\right)}^2={\left(m^2\right)}^2-2.m^2.1+1^2+4m^2\\ =m^4+2m^2+1\\ \therefore {\left(m^2+1\right)}^2={\left(m^2-1\right)}^2+{\left(2m\right)}^2\end{array}$

C,D are also same as the above process

But they are not form Pythagorean triplet