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Write a Pythagorean triplet whose one member is. (i) 6 (ii) 14 (iii) 16 (iv) 18

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Write a Pythagorean triplet whose one member is.

(i) 6

(ii) 14

(iii) 16

(iv) 18

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For any natural number m, we know that 2m, m2–1, m2+1 is a Pythagorean triplet.

(i) 2m = 6

⇒ m = 6/2 = 3

m2–1 = 32 – 1 = 9 – 1 = 8

m2 + 1 = 32 + 1 = 9 + 1 = 10

∴ (6, 8, 10) is a Pythagorean triplet.

(ii) 2m = 14

⇒ m = 14/2 = 7

m2–1= 72 – 1 = 49 – 1 = 48

m2+1 = 72 + 1 = 49 + 1 = 50

∴ (14, 48, 50) is not a Pythagorean triplet.

(iii) 2m = 16

⇒ m = 16/2 = 8

m2–1 = 82 – 1 = 64 – 1 = 63

m2+ 1 = 82 + 1 = 64 + 1 = 65

∴ (16, 63, 65) is a Pythagorean triplet.

(iv) 2m = 18

⇒ m = 18/2 = 9

m2–1 = 92 – 1 = 81 – 1 = 80

m2+1 = 92 + 1 = 81 + 1 = 82

∴ (18, 80, 82) is a Pythagorean triplet.

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