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Write a Pythagorean triplet whose one member is

(i) 6 (ii) 14

(iii) 16 (iv) 18

For any natural number *m* > 1, 2*m*, *m*^{2} - 1, *m*^{2}+ 1 forms a Pythagorean triplet

(i) If we consider,*m*^{2} + 1 = 6, then *m*^{2} = 5

The value of *m* will not be an integer

Now,

If we consider,*m*^{2} - 1 = 6, then *m*^{2} = 7

Again the value of *m* is not an integer

Let us take 2*m* = 6

*So, m* = 3

Hence, the Pythagorean triplets are:

2 x 3, 3^{2} - 1, 3^{2}+ 1 or 6, 8, and 10

(ii) If we consider*m*^{2} + 1 = 14, then *m*^{2} = 13

The value of *m* will not be an integer

If we consider *m*^{2} - 1 = 14, then *m*^{2} = 15

Again the value of *m* is not an integer

Let 2*m* = 14

*m* = 7

Hence, *m*^{2} - 1 = 49 - 1 = 48 and *m*^{2} + 1 = 49 + 1 = 50

Hence, the required Pythagorean triplet is 14, 48, and 50

(iii) If we consider*m*^{2} + 1 = 16, then *m*^{2} = 15

The value of *m* will not be an integer

If we consider*m*^{2} - 1= 16, then *m*^{2} = 17

Again the value of *m* is not an integer

Now, let 2*m* = 16

*m* = 8

Therefore, *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 consider*m*^{2} + 1 = 18,

*m*^{2} = 17

Then, the value of *m* will not be an integer

If we consider*m ^{2}* - 1 = 18, then

Again the value of *m* is not an integer

So, let 2*m* =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

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