Write a Pythagorean triplet whose one member is

(i) 6 (ii) 14


(iii) 16 (iv) 18

For any natural number m > 1, 2m, m2 - 1, m2+ 1 forms a Pythagorean triplet

(i) If we consider,m2 + 1 = 6, then m2 = 5


The value of m will not be an integer


Now,


If we consider,m2 - 1 = 6, then m2 = 7


Again the value of m is not an integer


Let us take 2m = 6


So, m = 3


Hence, the Pythagorean triplets are:


2 x 3, 32 - 1, 32+ 1 or 6, 8, and 10


(ii) If we considerm2 + 1 = 14, then m2 = 13


The value of m will not be an integer


If we consider m2 - 1 = 14, then m2 = 15


Again the value of m is not an integer


Let 2m = 14


m = 7


Hence, m2 - 1 = 49 - 1 = 48 and m2 + 1 = 49 + 1 = 50


Hence, the required Pythagorean triplet is 14, 48, and 50


(iii) If we considerm2 + 1 = 16, then m2 = 15


The value of m will not be an integer


If we considerm2 - 1= 16, then m2 = 17


Again the value of m is not an integer


Now, let 2m = 16


m = 8


Therefore, m2 - 1 = 64 - 1 = 63 and m2 + 1 = 64 + 1 = 65


Therefore, the Pythagorean triplet is 16, 63, and 65


(iv) If we considerm2 + 1 = 18,


m2 = 17


Then, the value of m will not be an integer


If we considerm2 - 1 = 18, then m2 = 19


Again the value of m is not an integer


So, let 2m =18


m = 9


Thus, m2 - 1 = 81 - 1 = 80 and m2 + 1 = 81 + 1 = 82


Therefore, the Pythagorean triplet is 18, 80, and 82


2