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