Question

Check whether $(6,8,10)$ is a Pythagorean triplet.

Answer 1

Solution:
$2m,m^2–1,m^2+1$ represent the Pythagorean triplet for any natural number $m$.
Let $2m=6\Rightarrow m=3$
$m^2–1=(3{)}^2–1=9–1=8$
and $m^2+1=(3{)}^2+1=9+1=10$
Hence $(6,8,10)$ is a Pythagorean triplet.

Alternative Method:
$(6{)}^2+(8{)}^2=36+64=100=(10{)}^2$
$\Rightarrow (6,8,10)$ is a Pythagorean triplet.