Among the following numbers, a Pythagorean triplet is

Three natural numbers  $m,n$ & $p$ are said to form a Pythagorean triplet $\left(m,n,p\right)$  if $m^2+n^2=p^2$ , for every natural number $m>1$ , we have $\left(2m,m^2-1,m^2+1\right)$  is a Pythagorean triplet.

Option ( A ) :  $\left(2,3,5\right)$

$2^2+3^2=5^2$

$4+9=25$

$13=25$  it is  impossible

$\therefore 13\ne 25$  it is not a Pythagorean triplet

Option ( B ) : $\left(5,7,9\right)$

$5^2+7^2=9^2$

$25+49=81$

$74=81$  it is  impossible

$\therefore 74\ne 81$   it is not a  Pythagorean triplet

Option ( C ) : $\left(6,9,11\right)$

$6^2+9^2={11}^2$

 $36+81=121$

  $117=121$  it is impossible

$\therefore 117\ne 121$  it is not a  Pythagorean triplet

Option ( D ): $\left(8,15,17\right)$

$8^2+{15}^2={17}^2$

$64+225=289$

$\therefore$     $289=289$  it is a  Pythagorean triplet