Question

Pythagorean triplets eg

• A. 6
• B. 14
• C. 16
• D. 18

Answer 1

Answer: (A), (B), (C), (D)

For any natural number m > 1, 2m, m^2- 1, m^2 + 1 forms a Pythagorean triplet. (i) If we take m^2+ 1 = 6, then m^2= 5 The value of m will not be an integer. If we take m^2- 1 = 6, then m^2= 7 Again the value of m is not an integer. Let 2m = 6 m = 3 Therefore, the Pythagorean triplets are 2 \times 3, 3^2- 1, 3^2+ 1 or 6, 8, and 10. (ii) If we take m^2+ 1 = 14, then m^2= 13 The value of m will not be an integer. If we take m^2- 1 = 14, then m^2= 15 Again the value of m is not an integer. Let 2m = 14 m = 7 Thus, m^2- 1 = 49 - 1 = 48 and m^2+ 1 = 49 + 1 = 50 Therefore, the required triplet is 14, 48, and 50. (iii) If we take m^2+ 1 = 16, then m^2= 15 The value of m will not be an integer. If we take m^2- 1= 16, then m^2= 17 Again the value of m is not an integer. Let 2m = 16 m = 8 Thus, m^2- 1 = 64 - 1 = 63 and m^2+ 1 = 64 + 1 = 65 Therefore, the Pythagorean triplet is 16, 63, and 65. (iv) If we take m^2+ 1 = 18, m^2= 17 The value of m will not be an integer. If we take m^2- 1 = 18, then m^2= 19 Again the value of m is not an integer. Let 2m =18 m = 9 Thus, m^2- 1 = 81 - 1 = 80 and m^2+ 1 = 81 + 1 = 82 Therefore, the Pythagorean triplet is 18, 80, and 82.