Example 2 Write a Pythagorean triplet whose smallest member is 8. We know 2m, π^2β1 and π^2+1 form a Pythagorean triplet.
We know 2m, π^πβπ and π^π+π form a Pythagorean triplet.
Given,
Smallest member of the triplet = 8.
Let 2m = 8
2m = 8
m = 8/2
m = 4
Let π^πβπ" = 8"
π^2 = 8 + 1
π^2 = 9
β΄ m = 3
Let π^π+π = 8
π^2 = 8 β 1
π^2 = 7
Since, 7 is not a square number,
β΄ π^2+1 β 8
It is not possible.
Therefore, m = 4 or m = 3
Finding Triplets for m = 3
1st number = 2m
2nd number = π^2β1
3rd number = π^2+1
β΄ The required triplet is 6, 8, 10
But 8 is not a smallest member of this triplet.
So, lets try for m = 4
Finding Triplets for m = 4
1st number = 2m
2nd number = π^2β1
3rd number = π^2+1
As 8 is the smallest member of this triplet.
β΄ The required triplet is 8, 15, 17

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.