Question

Write a Pythagorean triplet whose one member is

(i) 6 (ii) 14

(iii) 16 (iv) 18

Answer 1

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

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

The value of m will not be an integer.

If we take m² − 1 = 6, then m² = 7

Again the value of m is not an integer.

Let 2m = 6

m = 3

Therefore, the Pythagorean triplets are 2 × 3, 3² − 1, 3² + 1 or 6, 8, and 10.

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

The value of m will not be an integer.

If we take m² − 1 = 14, then m² = 15

Again the value of m is not an integer.

Let 2m = 14

m = 7

Thus, m² − 1 = 49 − 1 = 48 and m² + 1 = 49 + 1 = 50

Therefore, the required triplet is 14, 48, and 50.

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

The value of m will not be an integer.

If we take m² − 1= 16, then m² = 17

Again the value of m is not an integer.

Let 2m = 16

m = 8

Thus, m² − 1 = 64 − 1 = 63 and m² + 1 = 64 + 1 = 65

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

(iv) If we take m² + 1 = 18,

m² = 17

The value of m will not be an integer.

If we take m² − 1 = 18, then m² = 19

Again the value of m is not an integer.

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.