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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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