Question

Write a pythagorean triplet whose smallest number is

(i)6 (ii)16.

Answer 1

For any natural number $n>1$
$2n,n^2-1,n^2+1$ forms a pythagoras triplet.
Here,the smallest number is $2n$
Given,
$2n=6$
$=>n=3$
$\therefore$ Other are
$n^2-1=3^2-1$
$=9-1$
$=8$
And,
$n^2+1=3^2+1$
$=9+1$
$=10$
Again,
$2n=16$
$=>n=8$
$\therefore$ Other are
$n^2-1=8^2-1$
$=64-1$
$=63$
And,
$n^2+1=8^2+1$
$=64+1$
$=65$
Hence, pythagorean triplet smallest number are $(6,8,10),(16,63,65)$