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