Question

Write a pythagorean ripltes whose smallest nember is 6

Answer 1

2m, m^2-1 and m^2+1 are triplet.
So, let's first consider 2m to be 6. That is,
2m= 6 (Member 1)
m=3.

m^2-1 (Member 2)
3^2-1
9-1
8

m^2+1(Member 3)
3^2+1
9+1
10.
The triplet is 6,8 and 10.
Hope it helps!